Concept of limits, Evaluation of algebric limits Questions in English

(D) To evaluate the limit $L = \lim _{x \rightarrow-\infty}\left(\sqrt{4 x^2-x}+2 x\right)$,we rationalize the expression.
Multiply and divide by the conjugate $\left(\sqrt{4 x^2-x}-2 x\right)$:
$L = \lim _{x \rightarrow-\infty} \frac{(\sqrt{4 x^2-x}+2 x)(\sqrt{4 x^2-x}-2 x)}{\sqrt{4 x^2-x}-2 x}$
$L = \lim _{x \rightarrow-\infty} \frac{4 x^2 - x - 4x^2}{\sqrt{4 x^2-x}-2 x} = \lim _{x \rightarrow-\infty} \frac{-x}{\sqrt{x^2(4 - \frac{1}{x})} - 2x}$
Since $x \rightarrow -\infty$,we have $\sqrt{x^2} = |x| = -x$:
$L = \lim _{x \rightarrow-\infty} \frac{-x}{-x\sqrt{4 - \frac{1}{x}} - 2x} = \lim _{x \rightarrow-\infty} \frac{-x}{-x(\sqrt{4 - \frac{1}{x}} + 2)}$
$L = \lim _{x \rightarrow-\infty} \frac{1}{\sqrt{4 - \frac{1}{x}} + 2} = \frac{1}{\sqrt{4-0} + 2} = \frac{1}{2+2} = \frac{1}{4}$.

(B) We have $x_n = (2^n + 3^n)^{\frac{1}{2n}}$.
Taking the limit as $n \to \infty$:
$\lim_{n \to \infty} x_n = \lim_{n \to \infty} (3^n ((\frac{2}{3})^n + 1))^{\frac{1}{2n}}$
$= \lim_{n \to \infty} (3^n)^{\frac{1}{2n}} \cdot ((\frac{2}{3})^n + 1)^{\frac{1}{2n}}$
$= \lim_{n \to \infty} 3^{\frac{1}{2}} \cdot ((\frac{2}{3})^n + 1)^{\frac{1}{2n}}$
Since $\lim_{n \to \infty} (\frac{2}{3})^n = 0$ and $\lim_{n \to \infty} \frac{1}{2n} = 0$,the expression becomes:
$= \sqrt{3} \cdot (0 + 1)^0 = \sqrt{3} \cdot 1 = \sqrt{3}$.

(C) $\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right) f(x) = \lim _{x \rightarrow \infty} \frac{\sin (1/x)}{1/x} \cdot f(x) = 1 \cdot M = M$. This is true.
$(B)$ Since $\sin(x)$ is a continuous function,$\lim _{x \rightarrow \infty} \sin(f(x)) = \sin(\lim _{x \rightarrow \infty} f(x)) = \sin M$. This is true.
$(C)$ $\lim _{x \rightarrow \infty} x \sin \left(e^{-x}\right) f(x) = \lim _{x \rightarrow \infty} (x e^{-x}) \cdot \frac{\sin (e^{-x})}{e^{-x}} \cdot f(x)$. Since $\lim _{x \rightarrow \infty} x e^{-x} = 0$,the limit is $0 \cdot 1 \cdot M = 0$. Thus,the statement that the limit is $M$ is false.
$(D)$ $\lim _{x \rightarrow \infty} \frac{\sin x}{x} f(x) = (\lim _{x \rightarrow \infty} \frac{\sin x}{x}) \cdot (\lim _{x \rightarrow \infty} f(x)) = 0 \cdot M = 0$. This is true.

(D) We are given $\lim_{x \rightarrow a}(\lfloor x-5 \rfloor - \lfloor 2x+2 \rfloor) = 0$.
This simplifies to $\lim_{x \rightarrow a}(\lfloor x \rfloor - 5 - \lfloor 2x \rfloor - 2) = 0$,which implies $\lim_{x \rightarrow a}(\lfloor x \rfloor - \lfloor 2x \rfloor) = 7$.
For the limit to exist and equal $7$,the left-hand limit and right-hand limit must be equal.
Let $a = I + f$,where $I$ is an integer and $f \in [0, 1)$.
If $f \in [0, 0.5)$,then $\lfloor a \rfloor = I$ and $\lfloor 2a \rfloor = 2I$. The condition becomes $I - 2I = 7 \Rightarrow I = -7$. Thus $a \in [-7, -6.5)$.
If $f \in [0.5, 1)$,then $\lfloor a \rfloor = I$ and $\lfloor 2a \rfloor = 2I + 1$. The condition becomes $I - (2I + 1) = 7$ $\Rightarrow -I = 8$ $\Rightarrow I = -8$. Thus $a \in [-7.5, -7)$.
Combining these,$a \in [-7.5, -6.5)$.

(C) The numerator is a sum of $n$ groups of the form $(3k-2) + (3k-1) - 3k = 3k-3$ for $k=1$ to $n$.
Sum $= \sum_{k=1}^{n} (3k-3) = 3 \sum_{k=1}^{n} (k-1) = 3 \frac{(n-1)n}{2} = \frac{3n^2-3n}{2}$.
Now,consider the limit: $\operatorname{Lim}_{n \rightarrow \infty} \frac{\frac{3n^2-3n}{2}}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$.
Divide numerator and denominator by $n^2$: $\operatorname{Lim}_{n}$ ${\rightarrow \infty} \frac{\frac{3}{2} - \frac{3}{2n}}{\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}}$.
As $n \rightarrow \infty$,the expression approaches $\frac{3/2}{\sqrt{2}-1}$.
Rationalizing the denominator: $\frac{3}{2(\sqrt{2}-1)} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{3(\sqrt{2}+1)}{2(2-1)} = \frac{3}{2}(\sqrt{2}+1)$.

(B) Let the given expression be $L = \lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3$.
As $x \rightarrow \infty$,we can factor out $x$ from the terms inside the parentheses:
Numerator: $(\sqrt{3x+1} \pm \sqrt{3x-1})^6 = x^3 (\sqrt{3+1/x} \pm \sqrt{3-1/x})^6$.
Denominator: $(x \pm \sqrt{x^2-1})^6 = x^6 (1 \pm \sqrt{1-1/x^2})^6$.
Substituting these into the limit:
$L = \lim _{x}$ ${\rightarrow \infty} x^3 \cdot \frac{x^3 [(\sqrt{3+1/x} + \sqrt{3-1/x})^6 + (\sqrt{3+1/x} - \sqrt{3-1/x})^6]}{x^6 [(1 + \sqrt{1-1/x^2})^6 + (1 - \sqrt{1-1/x^2})^6]}$.
As $x \rightarrow \infty$,$1/x \rightarrow 0$ and $1/x^2 \rightarrow 0$.
$L = \frac{(\sqrt{3} + \sqrt{3})^6 + (\sqrt{3} - \sqrt{3})^6}{(1 + 1)^6 + (1 - 1)^6} = \frac{(2\sqrt{3})^6 + 0}{2^6 + 0} = \frac{64 \cdot 27}{64} = 27$.

(A) Let the given expression be $S_n = \sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$.
Rationalizing each term,we get $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$.
Thus,$S_n = \frac{1}{d} \sum_{k=1}^{n-1} (\sqrt{a_{k+1}} - \sqrt{a_k}) = \frac{\sqrt{a_n} - \sqrt{a_1}}{d}$.
The limit becomes $\lim_{n \rightarrow \infty} \sqrt{\frac{d}{n}} \cdot \frac{\sqrt{a_n} - \sqrt{a_1}}{d} = \lim_{n \rightarrow \infty} \frac{\sqrt{a_n} - \sqrt{a_1}}{\sqrt{nd}}$.
Since $a_n = a_1 + (n-1)d$,as $n \rightarrow \infty$,$a_n \approx nd$.
Therefore,the limit is $\lim_{n \rightarrow \infty} \frac{\sqrt{nd} - \sqrt{a_1}}{\sqrt{nd}} = \lim_{n \rightarrow \infty} \left( 1 - \frac{\sqrt{a_1}}{\sqrt{nd}} \right) = 1$.

(B) For $a$: $a = \lim_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} \times \frac{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}} = \lim_{x \rightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})} = \lim_{x \rightarrow 0} \frac{x^4}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$.
For $b$: $b = \lim_{x \rightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}} = \lim_{x \rightarrow 0} \frac{(1-\cos^2 x)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} = \lim_{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x} = \lim_{x \rightarrow 0} (1+\cos x)(\sqrt{2}+\sqrt{1+\cos x}) = (1+1)(\sqrt{2}+\sqrt{2}) = 2 \times 2\sqrt{2} = 4\sqrt{2}$.
Calculating $ab^3$: $ab^3 = \frac{1}{4\sqrt{2}} \times (4\sqrt{2})^3 = \frac{1}{4\sqrt{2}} \times 64 \times 2\sqrt{2} = \frac{128\sqrt{2}}{4\sqrt{2}} = 32$.

(A) For the right-hand limit $(R)$:
$\lim _{x \rightarrow 0^{+}} f(x) = \lim _{h \rightarrow 0^{+}} f(h) = \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}(1-h^2) \sin ^{-1}(1-h)}{h(1-h^2)}$
$= \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}(1-h^2)}{h} \cdot \frac{\sin ^{-1}(1)}{1} = \frac{\pi}{2} \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}(1-h^2)}{h}$.
Let $\cos ^{-1}(1-h^2) = \theta$,then $\cos \theta = 1-h^2$,so $h = \sqrt{1-\cos \theta} = \sqrt{2} \sin(\theta/2)$.
As $h \rightarrow 0^{+}$,$\theta \rightarrow 0^{+}$.
$R = \frac{\pi}{2} \lim _{\theta \rightarrow 0^{+}} \frac{\theta}{\sqrt{2} \sin(\theta/2)} = \frac{\pi}{2} \cdot \frac{1}{\sqrt{2} \cdot (1/2)} = \frac{\pi}{\sqrt{2}}$.
Thus,$R^2 = \frac{\pi^2}{2}$.
For the left-hand limit $(L)$:
$\lim _{x \rightarrow 0^{-}} f(x) = \lim _{h \rightarrow 0^{+}} f(-h) = \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}(1-(1-h)^2) \sin ^{-1}(1-(1-h))}{(1-h)-(1-h)^3} = \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}(2h-h^2) \sin ^{-1}(h)}{(1-h)(1-(1-h)^2)} = \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}(2h-h^2) \sin ^{-1}(h)}{(1-h)(2h-h^2)} = \lim _{h \rightarrow 0^{+}} \frac{\cos ^{-1}(0) \cdot h}{1 \cdot 2h} = \frac{\pi/2}{2} = \frac{\pi}{4}$.
Thus,$L^2 = \frac{\pi^2}{16}$.
Finally,$\frac{32}{\pi^2} (L^2 + R^2) = \frac{32}{\pi^2} \left(\frac{\pi^2}{16} + \frac{\pi^2}{2}\right) = \frac{32}{\pi^2} \left(\frac{\pi^2 + 8\pi^2}{16}\right) = 2 \cdot 9 = 18$.

(B) Let the numerator be $N = \sum_{r=1}^{n-1} (r^2-r)(n-r) = \sum_{r=1}^{n-1} (-r^3 + r^2(n+1) - nr)$.
Using standard summation formulas:
$N = -\left[\frac{(n-1)n}{2}\right]^2 + (n+1)\frac{(n-1)n(2n-1)}{6} - n\frac{(n-1)n}{2}$.
Let the denominator be $D = \sum_{r=1}^n r^3 - \sum_{r=1}^n r^2 = \left[\frac{n(n+1)}{2}\right]^2 - \frac{n(n+1)(2n+1)}{6}$.
As $n \rightarrow \infty$,the leading term of $N$ is $-\frac{n^4}{4} + \frac{2n^4}{6} = \frac{n^4}{12}$.
The leading term of $D$ is $\frac{n^4}{4}$.
Thus,$\lim _{n \rightarrow \infty} \frac{N}{D} = \frac{n^4/12}{n^4/4} = \frac{4}{12} = \frac{1}{3}$.

(B) Let $L = \lim _{x \rightarrow 0} 2\left(\frac{1-\prod_{k=1}^{10} (\cos kx)^{1/k}}{x^2}\right)$.
Using the expansion $\cos kx \approx 1 - \frac{(kx)^2}{2}$,we have $(\cos kx)^{1/k} \approx (1 - \frac{k^2x^2}{2})^{1/k} \approx 1 - \frac{1}{k} \cdot \frac{k^2x^2}{2} = 1 - \frac{kx^2}{2}$.
Thus,the product $\prod_{k=1}^{10} (\cos kx)^{1/k} \approx \prod_{k=1}^{10} (1 - \frac{kx^2}{2}) \approx 1 - \sum_{k=1}^{10} \frac{kx^2}{2}$.
Substituting this into the limit:
$L = \lim _{x \rightarrow 0} 2\left(\frac{1 - (1 - \sum_{k=1}^{10} \frac{kx^2}{2})}{x^2}\right)$
$L = \lim _{x \rightarrow 0} 2\left(\frac{\sum_{k=1}^{10} \frac{kx^2}{2}}{x^2}\right) = \sum_{k=1}^{10} k$.
$L = 1 + 2 + 3 + \ldots + 10 = \frac{10 \times 11}{2} = 55$.

(B) First,evaluate $\alpha$:
$\alpha = \lim_{x \rightarrow 0^{+}} \frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}}$.
Let $u = \sqrt{\tan x} - \sqrt{x}$. As $x \rightarrow 0^{+}$,$u \rightarrow 0$.
Using the limit $\lim_{u \rightarrow 0} \frac{e^{u+\sqrt{x}} - e^{\sqrt{x}}}{u} = e^{\sqrt{x}} \lim_{u \rightarrow 0} \frac{e^u - 1}{u} = e^{\sqrt{x}} \cdot 1$.
At $x=0$,$\alpha = e^0 = 1$.
Next,evaluate $\beta$:
$\beta = \lim_{x \rightarrow 0} (1 + \sin x)^{\frac{1}{2} \cot x} = e^{\lim_{x \rightarrow 0} \frac{1}{2} \cot x \cdot \sin x} = e^{\lim_{x \rightarrow 0} \frac{1}{2} \cos x} = e^{1/2} = \sqrt{e}$.
The quadratic equation is $ax^2 + bx - \sqrt{e} = 0$.
Since $\alpha = 1$ and $\beta = \sqrt{e}$ are roots,the sum of roots $\alpha + \beta = 1 + \sqrt{e} = -b/a$ and the product of roots $\alpha \beta = 1 \cdot \sqrt{e} = -\sqrt{e}/a$.
From the product,$\sqrt{e} = -\sqrt{e}/a \implies a = -1$.
From the sum,$1 + \sqrt{e} = -b/(-1) = b$.
Thus,$a = -1$ and $b = 1 + \sqrt{e}$.
Then $a + b = -1 + 1 + \sqrt{e} = \sqrt{e}$.
Finally,$12 \log_e(a + b) = 12 \log_e(\sqrt{e}) = 12 \times \frac{1}{2} = 6$.

(D) We use the standard limit formula $\lim_{x \rightarrow 0} (1 + f(x))^{\frac{1}{g(x)}} = e^{\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}}$.
Given the expression $\lim_{x \rightarrow 0} [1 + x \ln(1 + b^2)]^{\frac{1}{x}}$,we have $f(x) = x \ln(1 + b^2)$ and $g(x) = x$.
Thus,the limit is $e^{\lim_{x \rightarrow 0} \frac{x \ln(1 + b^2)}{x}} = e^{\ln(1 + b^2)} = 1 + b^2$.
Equating this to the given expression: $1 + b^2 = 2b \sin^2 \theta$.
Rearranging for $\sin^2 \theta$,we get $\sin^2 \theta = \frac{1 + b^2}{2b} = \frac{1}{2} (b + \frac{1}{b})$.
Since $b > 0$,by the $AM-GM$ inequality,$b + \frac{1}{b} \geq 2$,which implies $\sin^2 \theta \geq 1$.
Since the maximum value of $\sin^2 \theta$ is $1$,we must have $\sin^2 \theta = 1$.
This implies $\sin \theta = \pm 1$,so $\theta = \pm \frac{\pi}{2}$.

(B) Let $L = \lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^a}$.
We know that $(1-x)^{\frac{1}{x}} = e^{\frac{1}{x} \ln(1-x)}$.
Using the Taylor series expansion for $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$,we have:
$\frac{1}{x} \ln(1-x) = -1 - \frac{x}{2} - \frac{x^2}{3} - \dots$
Thus,$(1-x)^{\frac{1}{x}} = e^{-1 - \frac{x}{2} - \frac{x^2}{3} - \dots} = e^{-1} \cdot e^{-\frac{x}{2} - \frac{x^2}{3} - \dots}$.
Using $e^u = 1 + u + \frac{u^2}{2!} + \dots$,we get:
$(1-x)^{\frac{1}{x}} = e^{-1} \left( 1 + (-\frac{x}{2} - \frac{x^2}{3} - \dots) + \frac{(-\frac{x}{2} - \dots)^2}{2} + \dots \right) = e^{-1} \left( 1 - \frac{x}{2} - \frac{x^2}{3} + \frac{x^2}{8} + \dots \right) = e^{-1} \left( 1 - \frac{x}{2} - \frac{5x^2}{24} + \dots \right)$.
Substituting this into the limit:
$L = \lim_{x \rightarrow 0^{+}} \frac{e^{-1} (1 - \frac{x}{2} - \frac{5x^2}{24} + \dots) - e^{-1}}{x^a} = e^{-1} \lim_{x \rightarrow 0^{+}} \frac{-\frac{x}{2} - \frac{5x^2}{24} - \dots}{x^a}$.
For the limit to be a nonzero real number,the power of $x$ in the numerator must match the denominator $x^a$. Thus,$a = 1$.

(B) We are given the limit: $\lim_{\alpha \rightarrow 0} \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} = -\frac{e}{2}$.
Factor out $e$ from the numerator: $\lim_{\alpha \rightarrow 0} \frac{e(e^{\cos(\alpha^n)-1} - 1)}{\alpha^m} = -\frac{e}{2}$.
Using the standard limit $\lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1$,let $x = \cos(\alpha^n) - 1$. As $\alpha \rightarrow 0$,$x \rightarrow 0$.
So,$\lim_{\alpha}$ ${\rightarrow 0} \frac{e(e^{\cos(\alpha^n)-1} - 1)}{\cos(\alpha^n) - 1} \cdot \frac{\cos(\alpha^n) - 1}{\alpha^m} = -\frac{e}{2}$.
This simplifies to $e \cdot 1 \cdot \lim_{\alpha \rightarrow 0} \frac{\cos(\alpha^n) - 1}{\alpha^m} = -\frac{e}{2}$.
Using the expansion $\cos(x) \approx 1 - \frac{x^2}{2}$,we have $\cos(\alpha^n) - 1 \approx -\frac{(\alpha^n)^2}{2} = -\frac{\alpha^{2n}}{2}$.
Substituting this: $\lim_{\alpha \rightarrow 0} \frac{-\alpha^{2n}}{2\alpha^m} = -\frac{1}{2}$.
For this limit to be a non-zero constant,the powers of $\alpha$ must match,so $2n = m$.
Thus,$\frac{m}{n} = 2$.

(A) We are given $\beta = \lim_{x \rightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$.
Using the standard limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$,we can rewrite the denominator as $x \sin^2 x = x^3 \left(\frac{\sin x}{x}\right)^2 \approx x^3$ as $x \rightarrow 0$.
Expanding the terms using Taylor series:
$e^{x^3} = 1 + x^3 + O(x^6)$
$(1 - x^3)^{1/3} = 1 - \frac{1}{3}x^3 + O(x^6)$
$(1 - x^2)^{1/2} - 1 = (1 - \frac{1}{2}x^2 + O(x^4)) - 1 = -\frac{1}{2}x^2 + O(x^4)$
Substituting these into the expression:
$\beta = \lim_{x \rightarrow 0} \frac{(1 + x^3) - (1 - \frac{1}{3}x^3) + (-\frac{1}{2}x^2) \cdot x}{x^3}$
$\beta = \lim_{x \rightarrow 0} \frac{1 + x^3 - 1 + \frac{1}{3}x^3 - \frac{1}{2}x^3}{x^3}$
$\beta = \lim_{x}$ ${\rightarrow 0} \frac{(1 + \frac{1}{3} - \frac{1}{2})x^3}{x^3} = 1 + \frac{1}{3} - \frac{1}{2} = \frac{6 + 2 - 3}{6} = \frac{5}{6}$.
Therefore,$6 \beta = 6 \times \frac{5}{6} = 5$.

(B, C) Given the limit: $\lim _{x \rightarrow \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin(1/x^2)}{x^{\alpha \beta}(\log_e(1+x))^\beta} = 0$.
Since $\sin(x^2)$ is bounded and $\sin(1/x^2) \approx 1/x^2$ as $x \rightarrow \infty$,and $\log_e(1+x) \approx \log_e x$ as $x \rightarrow \infty$,the expression behaves as:
$\lim _{x \rightarrow \infty} \frac{(\log_e x)^\alpha \cdot (1/x^2)}{x^{\alpha \beta} \cdot (\log_e x)^\beta} = \lim _{x \rightarrow \infty} \frac{(\log_e x)^{\alpha-\beta}}{x^{\alpha \beta+2}} = 0$.
For this limit to be $0$,the exponent of $x$ in the denominator must be positive,i.e.,$\alpha \beta + 2 > 0$,which implies $\alpha \beta > -2$.
Checking the options:
$(A) (-1, 3) \implies (-1)(3) = -3 \ngtr -2$ (Incorrect).
$(B) (-1, 1) \implies (-1)(1) = -1 > -2$ (Correct).
$(C) (1, -1) \implies (1)(-1) = -1 > -2$ (Correct).
$(D) (1, -2) \implies (1)(-2) = -2 \ngtr -2$ (Incorrect).
Thus,$(B)$ and $(C)$ are correct.

(A) Given $\alpha = \lim _{x \rightarrow \infty} \left( \left( \frac{e}{1-e} \right) \left( \frac{1}{e} - \frac{x}{1+x} \right) \right)^x$,which is in the $1^{\infty}$ form.
We know that $\lim _{x \rightarrow \infty} f(x)^{g(x)} = e^{\lim _{x \rightarrow \infty} g(x)(f(x)-1)}$.
Let $L = \lim _{x \rightarrow \infty} x \left( \left( \frac{e}{1-e} \right) \left( \frac{1}{e} - \frac{x}{1+x} \right) - 1 \right)$.
Simplifying the expression inside the limit:
$\left( \frac{e}{1-e} \right) \left( \frac{1}{e} - \frac{x}{1+x} \right) = \frac{1}{1-e} - \frac{ex}{(1-e)(1+x)} = \frac{1+x-ex}{(1-e)(1+x)}$.
Now,$L = \lim _{x \rightarrow \infty} x \left( \frac{1+x-ex}{(1-e)(1+x)} - 1 \right) = \lim _{x \rightarrow \infty} x \left( \frac{1+x-ex - (1-e)(1+x)}{(1-e)(1+x)} \right)$.
$L = \lim _{x \rightarrow \infty} x \left( \frac{1+x-ex - (1+x-e-ex)}{(1-e)(1+x)} \right) = \lim _{x \rightarrow \infty} x \left( \frac{e}{(1-e)(1+x)} \right) = \frac{e}{1-e}$.
Thus,$\alpha = e^{\frac{e}{1-e}}$,so $\log _e \alpha = \frac{e}{1-e}$.
The required value is $\frac{\log _e \alpha}{1+\log _e \alpha} = \frac{\frac{e}{1-e}}{1 + \frac{e}{1-e}} = \frac{e}{1-e+e} = e$.

(D) Given limit: $L = \lim _{x \rightarrow \infty} \frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)(3x+2)^{x/2}}$
$= \lim _{x \rightarrow \infty} \frac{x^2(2-3/x+5/x^2)}{x^2(3+5/x+4/x^2)} \cdot \left( \frac{3x-1}{3x+2} \right)^{x/2}$
$= \frac{2}{3} \cdot \lim _{x \rightarrow \infty} \left( \frac{1-1/(3x)}{1+2/(3x)} \right)^{x/2}$
$= \frac{2}{3} \cdot \frac{\lim _{x \rightarrow \infty} (1-1/(3x))^{x/2}}{\lim _{x \rightarrow \infty} (1+2/(3x))^{x/2}}$
$= \frac{2}{3} \cdot \frac{e^{\lim _{x \rightarrow \infty} (x/2)(-1/(3x))}}{e^{\lim _{x \rightarrow \infty} (x/2)(2/(3x)))}}$
$= \frac{2}{3} \cdot \frac{e^{-1/6}}{e^{1/3}} = \frac{2}{3} e^{-1/6 - 1/3} = \frac{2}{3} e^{-1/2} = \frac{2}{3\sqrt{e}}$

(D) Let $L = \lim _{x \rightarrow 0} \frac{\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}}{\sin x}$.
Rationalizing the numerator,we get:
$L = \lim _{x}$ ${\rightarrow 0} \frac{(2 \cos ^2 x+3 \cos x) - (\cos ^2 x+\sin x+4)}{\sin x (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$
$L = \lim _{x}$ ${\rightarrow 0} \frac{\cos ^2 x+3 \cos x - \sin x - 4}{\sin x (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$
Since $\cos ^2 x + 3 \cos x - 4 = (\cos x - 1)(\cos x + 4)$,we have:
$L = \lim _{x}$ ${\rightarrow 0} \frac{(\cos x - 1)(\cos x + 4) - \sin x}{\sin x (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$
Using $\cos x - 1 = -2 \sin ^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:
$L = \lim _{x}$ ${\rightarrow 0} \frac{-2 \sin ^2 \frac{x}{2}(\cos x + 4) - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2} (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$
Dividing numerator and denominator by $2 \sin \frac{x}{2}$:
$L = \lim _{x}$ ${\rightarrow 0} \frac{-\sin \frac{x}{2}(\cos x + 4) - \cos \frac{x}{2}}{\cos \frac{x}{2} (\sqrt{2 \cos ^2 x+3 \cos x}+\sqrt{\cos ^2 x+\sin x+4})}$
As $x \rightarrow 0$,$\sin \frac{x}{2} \rightarrow 0$,$\cos \frac{x}{2} \rightarrow 1$,$\cos x \rightarrow 1$:
$L = \frac{-(0)(5) - 1}{1 (\sqrt{2+3} + \sqrt{1+0+4})} = \frac{-1}{\sqrt{5} + \sqrt{5}} = -\frac{1}{2 \sqrt{5}}$.

(C) We know that $\lim_{x \rightarrow 0^{+}} x[\frac{k}{x}] = k$ for any constant $k > 0$.
Applying this to the given limit:
$\lim_{x \rightarrow 0^{+}} x \sum_{k=1}^{p} [\frac{k}{x}] = \sum_{k=1}^{p} k = \frac{p(p+1)}{2}$.
Similarly,$\lim_{x \rightarrow 0^{+}} x^2 [\frac{k^2}{x^2}] = k^2$.
So,$\lim_{x}$ ${\rightarrow 0^{+}} x^2 \sum_{k=1}^{9} [\frac{k^2}{x^2}] = \sum_{k=1}^{9} k^2 = \frac{9(9+1)(2 \times 9 + 1)}{6} = \frac{9 \times 10 \times 19}{6} = 285$.
The inequality becomes:
$\frac{p(p+1)}{2} - 285 \geq 1$
$\frac{p(p+1)}{2} \geq 286$
$p(p+1) \geq 572$.
For $p=23$,$23 \times 24 = 552 < 572$.
For $p=24$,$24 \times 25 = 600 \geq 572$.
Thus,the least natural value of $p$ is $24$.

(C) Let $P = \lim_{x \rightarrow 0} \left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}$.
Since the form is $1^{\infty}$,we use the formula $\lim_{x \rightarrow a} f(x)^{g(x)} = e^{\lim_{x \rightarrow a} (f(x)-1)g(x)}$.
$P = e^{\lim_{x \rightarrow 0} \left(\frac{\tan x}{x} - 1\right) \frac{1}{x^2}} = e^{\lim_{x \rightarrow 0} \frac{\tan x - x}{x^3}}$.
Using the Taylor series expansion $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$,we get:
$P = e^{\lim_{x \rightarrow 0} \frac{(x + \frac{x^3}{3} + \dots) - x}{x^3}} = e^{\lim_{x \rightarrow 0} \frac{x^3/3}{x^3}} = e^{1/3}$.
Thus,$\log_e p = \log_e (e^{1/3}) = \frac{1}{3}$.
Therefore,$96 \log_e p = 96 \times \frac{1}{3} = 32$.

(C) We use the standard limits: $\lim_{u \rightarrow 0} \frac{\tan u}{u} = 1$,$\lim_{u \rightarrow 0} \frac{\ln(1+u)}{u} = 1$,$\lim_{u \rightarrow 0} \frac{\tan^{-1} u}{u} = 1$,and $\lim_{u \rightarrow 0} \frac{e^u-1}{u} = 1$.
Given expression: $L = \lim _{x}$ ${\rightarrow 0^{+}} \frac{\tan \left(5 x^{1 / 3}\right)}{5 x^{1 / 3}} \cdot \frac{5 x^{1 / 3}}{\left(\tan ^{-1} 3 x^{1 / 2}\right)^2} \cdot \frac{\ln(1+3 x^2)}{3 x^2} \cdot \frac{3 x^2}{e^{5 x^{4 / 3}}-1}$.
Note that $\left(\tan ^{-1} 3 x^{1 / 2}\right)^2 = (3 x^{1 / 2})^2 \cdot \left(\frac{\tan ^{-1} 3 x^{1 / 2}}{3 x^{1 / 2}}\right)^2 = 9x \cdot (1)^2 = 9x$.
Also,$e^{5 x^{4 / 3}}-1 = 5 x^{4 / 3} \cdot \frac{e^{5 x^{4 / 3}}-1}{5 x^{4 / 3}} = 5 x^{4 / 3} \cdot (1)$.
Substituting these: $L = 1 \cdot \frac{5 x^{1 / 3}}{9x} \cdot 1 \cdot \frac{3 x^2}{5 x^{4 / 3}} = \frac{15 x^{7 / 3}}{45 x^{7 / 3}} = \frac{15}{45} = \frac{1}{3}$.

(D) For Statement $I$:
Using Taylor series expansions:
$\tan ^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$
$\log _e \sqrt{\frac{1+x}{1-x}} = \frac{1}{2} [\ln(1+x) - \ln(1-x)] = \frac{1}{2} [(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}) - (-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5})] = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$
Substituting these into the limit:
$\lim _{x \rightarrow 0} \frac{(x - \frac{x^3}{3} + \frac{x^5}{5}) + (x + \frac{x^3}{3} + \frac{x^5}{5}) - 2x}{x^5} = \lim _{x \rightarrow 0} \frac{2x + \frac{2x^5}{5} - 2x}{x^5} = \frac{2}{5}$.
Thus,Statement $I$ is true.
For Statement $II$:
Let $L = \lim _{x \rightarrow 1} x^{\frac{2}{1-x}}$. This is a $1^\infty$ form.
$L = e^{\lim _{x \rightarrow 1} (x-1) \cdot \frac{2}{1-x}} = e^{\lim _{x \rightarrow 1} -2} = e^{-2} = \frac{1}{e^2}$.
Thus,Statement $II$ is true.

(A) Given $f(x) = 5 - |x - 2|$. Since $|x - 2| \geq 0$,the maximum value of $f(x)$ is $5$,which occurs when $|x - 2| = 0$,i.e.,$x = 2$. Thus,$\alpha = 2$.
Given $g(x) = |x + 1|$. Since $|x + 1| \geq 0$,the minimum value of $g(x)$ is $0$,which occurs when $|x + 1| = 0$,i.e.,$x = -1$. Thus,$\beta = -1$.
We need to evaluate $\lim _{x \rightarrow -\alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$.
Since $-\alpha \beta = -(2)(-1) = 2$,the limit is $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 2)(x - 3)}{(x - 2)(x - 4)}$.
Canceling the common factor $(x - 2)$,we get $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 3)}{x - 4}$.
Substituting $x = 2$,we get $\frac{(2 - 1)(2 - 3)}{2 - 4} = \frac{(1)(-1)}{-2} = \frac{-1}{-2} = \frac{1}{2}$.

(B) To find the limit $\lim _{x \rightarrow 0} \frac{|x|}{|x|+x^2}$,we evaluate the left-hand limit and the right-hand limit.
For the left-hand limit $(x \rightarrow 0^-)$,we have $|x| = -x$. Thus,the expression becomes $\frac{-x}{-x+x^2} = \frac{-x}{x(x-1)} = \frac{-1}{x-1}$. As $x \rightarrow 0^-$,this approaches $\frac{-1}{0-1} = 1$.
For the right-hand limit $(x \rightarrow 0^+)$,we have $|x| = x$. Thus,the expression becomes $\frac{x}{x+x^2} = \frac{x}{x(1+x)} = \frac{1}{1+x}$. As $x \rightarrow 0^+$,this approaches $\frac{1}{1+0} = 1$.
Since the left-hand limit equals the right-hand limit,the limit exists and is equal to $1$.

(C) Given the limit: $\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$.
For $x \rightarrow 0^{-}$,we have $[x] = -1$ and $|x| = -x$.
Substituting these values into the expression:
$\lim _{x \rightarrow 0^{-}} \frac{x(-1 + (-x)) \sin(-1)}{-x}$
$= \lim _{x \rightarrow 0^{-}} \frac{x(-1 - x)(-\sin 1)}{-x}$
$= \lim _{x \rightarrow 0^{-}} (1 + x) \sin 1$
As $x \rightarrow 0$,this becomes $(1 + 0) \sin 1 = \sin 1$.

(B) To find the limit $\lim _{x \rightarrow 0} \frac{2x}{|x|+x^2}$,we evaluate the Left-Hand Limit $(LHL)$ and Right-Hand Limit $(RHL)$.
$LHL = \lim _{x \rightarrow 0^{-}} \frac{2x}{-x+x^2} = \lim _{x \rightarrow 0^{-}} \frac{2}{-1+x} = \frac{2}{-1} = -2$.
$RHL = \lim _{x \rightarrow 0^{+}} \frac{2x}{x+x^2} = \lim _{x \rightarrow 0^{+}} \frac{2}{1+x} = \frac{2}{1} = 2$.
Since $LHL \neq RHL$,the limit does not exist.

(A) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{e^{x^4}-1}{e^{x^4}+1}$,we divide the numerator and the denominator by $e^{x^4}$.
$\lim _{x \rightarrow \infty} \frac{e^{x^4}(1 - \frac{1}{e^{x^4}})}{e^{x^4}(1 + \frac{1}{e^{x^4}})} = \lim _{x \rightarrow \infty} \frac{1 - \frac{1}{e^{x^4}}}{1 + \frac{1}{e^{x^4}}}$.
As $x \rightarrow \infty$,$x^4 \rightarrow \infty$,so $e^{x^4} \rightarrow \infty$.
Therefore,$\frac{1}{e^{x^4}} \rightarrow 0$.
Substituting this into the expression,we get $\frac{1 - 0}{1 + 0} = 1$.

(B) Let $L = \lim _{x \rightarrow 0} \frac{63^x-9^x-7^x+1}{\sqrt{2}-\sqrt{1+\cos x}}$.
Factor the numerator: $63^x-9^x-7^x+1 = 9^x(7^x-1) - 1(7^x-1) = (9^x-1)(7^x-1)$.
Simplify the denominator: $\sqrt{2}-\sqrt{1+\cos x} = \sqrt{2}-\sqrt{2\cos^2(x/2)} = \sqrt{2}(1-\cos(x/2))$.
Using the identity $1-\cos \theta = 2\sin^2(\theta/2)$,we have $\sqrt{2}(2\sin^2(x/4)) = 2\sqrt{2}\sin^2(x/4)$.
Thus,$L = \lim _{x \rightarrow 0} \frac{(9^x-1)(7^x-1)}{2\sqrt{2}\sin^2(x/4)}$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \ln a$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
$L = \lim _{x}$ ${\rightarrow 0} \frac{(\frac{9^x-1}{x})(\frac{7^x-1}{x})x^2}{2\sqrt{2}(\frac{\sin(x/4)}{x/4})^2 (x/4)^2} = \frac{\ln 9 \cdot \ln 7}{2\sqrt{2} \cdot (1/16)} = \frac{16 \ln 9 \cdot \ln 7}{2\sqrt{2}} = 4\sqrt{2} \ln 9 \ln 7$.

(B) We are given the limit $L = \lim _{x \rightarrow \infty} \frac{\sum_{k=1}^{100} (2x+k)^{50}}{(2x)^{50} + 10^{50}}$.
Divide the numerator and denominator by $(2x)^{50}$:
$L = \lim _{x}$ ${\rightarrow \infty} \frac{\sum_{k=1}^{100} \left(\frac{2x+k}{2x}\right)^{50}}{1 + \frac{10^{50}}{(2x)^{50}}}$.
As $x \rightarrow \infty$,the term $\frac{k}{2x} \rightarrow 0$ for each $k \in \{1, 2, \dots, 100\}$.
Thus,$\left(1 + \frac{k}{2x}\right)^{50} \rightarrow 1^{50} = 1$.
The numerator becomes $\sum_{k=1}^{100} 1 = 100$.
The denominator becomes $1 + 0 = 1$.
Therefore,$L = \frac{100}{1} = 100$.

(C) We evaluate the limit using the standard series expansions for $x \rightarrow 0$:
$e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \dots$
$\cos 3x = 1 - \frac{(3x)^2}{2} + \dots = 1 - \frac{9x^2}{2} + \dots$
$\sin x = x + \dots$
$\log(1+2x) = 2x - \frac{(2x)^2}{2} + \dots = 2x - 2x^2 + \dots$
Substituting these into the expression:
$\lim _{x \rightarrow 0} \frac{(1 + x^2 + \dots) - (1 - \frac{9x^2}{2} + \dots)}{(x + \dots)(2x - \dots)} = \lim _{x \rightarrow 0} \frac{x^2 + \frac{9x^2}{2}}{2x^2} = \lim _{x \rightarrow 0} \frac{\frac{11x^2}{2}}{2x^2} = \frac{11}{4}$.

(D) To find the limit,we evaluate the left-hand limit $(L.H.L.)$ and the right-hand limit $(R.H.L.)$ at $x = 0$.
$L.H.L. = \lim _{x \rightarrow 0^{-}} \frac{x}{|x|+x^2} = \lim _{x \rightarrow 0^{-}} \frac{x}{-x+x^2} = \lim _{x \rightarrow 0^{-}} \frac{1}{-1+x} = \frac{1}{-1} = -1$.
$R.H.L. = \lim _{x \rightarrow 0^{+}} \frac{x}{|x|+x^2} = \lim _{x \rightarrow 0^{+}} \frac{x}{x+x^2} = \lim _{x \rightarrow 0^{+}} \frac{1}{1+x} = \frac{1}{1} = 1$.
Since $L.H.L. \neq R.H.L.$,the limit does not exist.

(A) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{9^x-4^x}{x(9^x+4^x)}$
Using the standard limit formula $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \log(a)$:
$= \lim _{x \rightarrow 0} \left( \frac{9^x-1 - (4^x-1)}{x} \right) \cdot \frac{1}{9^x+4^x}$
$= \left( \lim _{x \rightarrow 0} \frac{9^x-1}{x} - \lim _{x \rightarrow 0} \frac{4^x-1}{x} \right) \cdot \lim _{x \rightarrow 0} \frac{1}{9^x+4^x}$
$= (\log(9) - \log(4)) \cdot \frac{1}{1+1}$
$= \log \left( \frac{9}{4} \right) \cdot \frac{1}{2}$
$= \log \left( \left( \frac{3}{2} \right)^2 \right) \cdot \frac{1}{2}$
$= 2 \log \left( \frac{3}{2} \right) \cdot \frac{1}{2} = \log \left( \frac{3}{2} \right)$

(C) Let $L = \lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)$.
Let $t = 5^{\frac{x}{2}}$. As $x \rightarrow 2$,$t \rightarrow 5^1 = 5$.
Then $5^x = t^2$ and $5^{3-x} = \frac{125}{5^x} = \frac{125}{t^2}$.
Substituting these into the limit:
$L = \lim _{t \rightarrow 5} \frac{t^2 + \frac{125}{t^2} - 30}{\frac{125}{t^2} - t} = \lim _{t \rightarrow 5} \frac{t^4 - 30t^2 + 125}{125 - t^3}$.
Factorizing the numerator: $t^4 - 30t^2 + 125 = (t^2 - 25)(t^2 - 5) = (t-5)(t+5)(t^2-5)$.
Factorizing the denominator: $125 - t^3 = (5-t)(25 + 5t + t^2) = -(t-5)(25 + 5t + t^2)$.
Thus,$L = \lim _{t \rightarrow 5} \frac{(t-5)(t+5)(t^2-5)}{-(t-5)(25 + 5t + t^2)} = \lim _{t \rightarrow 5} \frac{-(t+5)(t^2-5)}{25 + 5t + t^2}$.
Substituting $t = 5$:
$L = \frac{-(5+5)(25-5)}{25 + 5(5) + 25} = \frac{-10 \times 20}{75} = \frac{-200}{75} = \frac{-8}{3}$.

(A) Let $f(x) = \frac{3^x + 3^{3-x} - 12}{3^{3-x} - 3^{x/2}}$.
Multiply the numerator and denominator by $3^x$:
$\lim _{x \rightarrow 2} \frac{3^{2x} - 12 \cdot 3^x + 27}{3^3 - 3^{3x/2}}$
Let $t = 3^{x/2}$. As $x \rightarrow 2$,$t \rightarrow 3$.
The expression becomes $\lim _{t \rightarrow 3} \frac{t^4 - 12t^2 + 27}{27 - t^3}$.
Factor the numerator: $t^4 - 12t^2 + 27 = (t^2 - 9)(t^2 - 3) = (t-3)(t+3)(t^2-3)$.
Factor the denominator: $27 - t^3 = (3-t)(9 + 3t + t^2) = -(t-3)(t^2 + 3t + 9)$.
$\lim _{t \rightarrow 3} \frac{(t-3)(t+3)(t^2-3)}{-(t-3)(t^2 + 3t + 9)} = \lim _{t \rightarrow 3} \frac{-(t+3)(t^2-3)}{t^2 + 3t + 9}$.
Substitute $t=3$: $\frac{-(3+3)(9-3)}{9+9+9} = \frac{-6 \cdot 6}{27} = \frac{-36}{27} = -\frac{4}{3}$.

(C) To evaluate the limit,we rationalize the numerator:
$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} \times \frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$
$= \lim _{y \rightarrow 0} \frac{1+\sqrt{1+y^4}-2}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})} = \lim _{y \rightarrow 0} \frac{\sqrt{1+y^4}-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})}$
Now,rationalize the remaining square root term:
$= \lim _{y \rightarrow 0} \frac{(\sqrt{1+y^4}-1)(\sqrt{1+y^4}+1)}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$
$= \lim _{y \rightarrow 0} \frac{1+y^4-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$
$= \lim _{y \rightarrow 0} \frac{y^4}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$
$= \frac{1}{(\sqrt{1+1}+\sqrt{2})(1+1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$

(C) We evaluate the limit $\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$ by rationalizing both the numerator and the denominator.
Multiplying by the conjugates:
$\lim _{x}$ ${\rightarrow a} \frac{(\sqrt{a+2 x}-\sqrt{3 x})(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})}$
Simplifying the terms using $(A-B)(A+B) = A^2 - B^2$:
$\lim _{x \rightarrow a} \frac{(a+2 x-3 x)(\sqrt{3 a+x}+2 \sqrt{x})}{(3 a+x-4 x)(\sqrt{a+2 x}+\sqrt{3 x})}$
$\lim _{x \rightarrow a} \frac{(a-x)(\sqrt{3 a+x}+2 \sqrt{x})}{3(a-x)(\sqrt{a+2 x}+\sqrt{3 x})}$
Canceling the common factor $(a-x)$:
$\lim _{x \rightarrow a} \frac{\sqrt{3 a+x}+2 \sqrt{x}}{3(\sqrt{a+2 x}+\sqrt{3 x})}$
Substituting $x = a$:
$\frac{\sqrt{4a} + 2\sqrt{a}}{3(\sqrt{3a} + \sqrt{3a})} = \frac{2\sqrt{a} + 2\sqrt{a}}{3(2\sqrt{3a})} = \frac{4\sqrt{a}}{6\sqrt{3a}} = \frac{2}{3\sqrt{3}}$

(C) We evaluate the limit: $\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$
Rationalizing the numerator and denominator:
$= \lim _{x}$ ${\rightarrow a} \frac{(\sqrt{a+2 x}-\sqrt{3 x})(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})}$
$= \lim _{x \rightarrow a} \frac{(a+2 x-3 x)(\sqrt{3 a+x}+2 \sqrt{x})}{(3 a+x-4 x)(\sqrt{a+2 x}+\sqrt{3 x})}$
$= \lim _{x \rightarrow a} \frac{(a-x)(\sqrt{3 a+x}+2 \sqrt{x})}{3(a-x)(\sqrt{a+2 x}+\sqrt{3 x})}$
Canceling $(a-x)$ and substituting $x = a$:
$= \frac{\sqrt{3 a+a}+2 \sqrt{a}}{3(\sqrt{a+2 a}+\sqrt{3 a})}$
$= \frac{2 \sqrt{a}+2 \sqrt{a}}{3(\sqrt{3 a}+\sqrt{3 a})}$
$= \frac{4 \sqrt{a}}{3(2 \sqrt{3 a})} = \frac{4 \sqrt{a}}{6 \sqrt{3} \sqrt{a}} = \frac{2}{3 \sqrt{3}}$

(C) We are given the limit: $\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$
First,factor the denominator of the second term: $x^3-3x^2+2x = x(x^2-3x+2) = x(x-2)(x-1)$
Now,substitute this back into the limit: $\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x(x-2)(x-1)}\right]$
Find a common denominator: $\lim _{x \rightarrow 2}\left[\frac{x(x-1)-2}{x(x-2)(x-1)}\right]$
Simplify the numerator: $x^2-x-2 = (x-2)(x+1)$
Substitute the simplified numerator: $\lim _{x \rightarrow 2}\left[\frac{(x-2)(x+1)}{x(x-2)(x-1)}\right]$
Cancel the common factor $(x-2)$: $\lim _{x \rightarrow 2}\frac{x+1}{x(x-1)}$
Evaluate the limit by substituting $x=2$: $\frac{2+1}{2(2-1)} = \frac{3}{2(1)} = \frac{3}{2}$

(A) Given $f(x) = 5 - |x - 2|$. The maximum value of $f(x)$ occurs when $|x - 2| = 0$,so $\alpha = 2$.
Given $g(x) = |x + 1|$. The minimum value of $g(x)$ occurs when $x + 1 = 0$,so $\beta = -1$.
The limit is to be evaluated at $x \rightarrow -\alpha \beta = - (2)(-1) = 2$.
We need to calculate $\lim _{x \rightarrow 2} \frac{(x - 1)(x^2 - 5x + 6)}{(x^2 - 6x + 8)}$.
Factorizing the expressions: $x^2 - 5x + 6 = (x - 2)(x - 3)$ and $x^2 - 6x + 8 = (x - 2)(x - 4)$.
Substituting these into the limit: $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 2)(x - 3)}{(x - 2)(x - 4)}$.
Canceling the common factor $(x - 2)$: $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 3)}{x - 4}$.
Evaluating the limit: $\frac{(2 - 1)(2 - 3)}{2 - 4} = \frac{(1)(-1)}{-2} = \frac{-1}{-2} = \frac{1}{2}$.

(D) To evaluate the limit $\lim _{n \rightarrow \infty} n\left(\sqrt{n^2+9}-n\right)$,we rationalize the expression by multiplying and dividing by the conjugate $\left(\sqrt{n^2+9}+n\right)$.
$\lim _{n \rightarrow \infty} n\left(\sqrt{n^2+9}-n\right) \times \frac{\sqrt{n^2+9}+n}{\sqrt{n^2+9}+n}$
$= \lim _{n \rightarrow \infty} \frac{n(n^2+9-n^2)}{\sqrt{n^2+9}+n}$
$= \lim _{n \rightarrow \infty} \frac{9n}{\sqrt{n^2+9}+n}$
Divide the numerator and denominator by $n$:
$= \lim _{n \rightarrow \infty} \frac{9}{\sqrt{\frac{n^2+9}{n^2}}+1}$
$= \lim _{n \rightarrow \infty} \frac{9}{\sqrt{1+\frac{9}{n^2}}+1}$
As $n \rightarrow \infty$,$\frac{9}{n^2} \rightarrow 0$,so the limit is $\frac{9}{\sqrt{1+0}+1} = \frac{9}{1+1} = \frac{9}{2}$.

(B) First,we calculate $m$:
$m = \lim_{x \rightarrow 0} \frac{x \log(1+2x)}{x \tan x} = \lim_{x \rightarrow 0} \frac{\log(1+2x)}{\tan x} = \lim_{x \rightarrow 0} \left( \frac{\log(1+2x)}{2x} \times \frac{x}{\tan x} \times 2 \right) = 1 \times 1 \times 2 = 2$.
Next,we calculate $n$:
$n = \lim_{x \rightarrow 0} \frac{\log x + \log(\frac{1+x}{x})}{x} = \lim_{x \rightarrow 0} \frac{\log(x \times \frac{1+x}{x})}{x} = \lim_{x \rightarrow 0} \frac{\log(1+x)}{x} = 1$.
The quadratic equation with roots $m=2$ and $n=1$ is given by $x^2 - (m+n)x + mn = 0$.
Substituting the values,we get $x^2 - (2+1)x + (2 \times 1) = 0$,which simplifies to $x^2 - 3x + 2 = 0$.

(C) To evaluate the limit $\lim _{x \rightarrow \infty}\left(\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right)^{\frac{4 x+3}{8 x-1}}$,we first find the limit of the base as $x \rightarrow \infty$:
$\lim _{x \rightarrow \infty} \frac{8 x^2+5 x+3}{2 x^2-7 x-5} = \lim _{x \rightarrow \infty} \frac{8 + \frac{5}{x} + \frac{3}{x^2}}{2 - \frac{7}{x} - \frac{5}{x^2}} = \frac{8+0+0}{2-0-0} = 4$.
Next,we find the limit of the exponent as $x \rightarrow \infty$:
$\lim _{x \rightarrow \infty} \frac{4 x+3}{8 x-1} = \lim _{x \rightarrow \infty} \frac{4 + \frac{3}{x}}{8 - \frac{1}{x}} = \frac{4+0}{8-0} = \frac{4}{8} = \frac{1}{2}$.
Therefore,the limit is $4^{\frac{1}{2}} = 2$.

(C) To evaluate the limit $\lim _{x \rightarrow \infty} (\sqrt{x^2+5x-7}-x)$,we rationalize the expression:
$\lim _{x \rightarrow \infty} \frac{(\sqrt{x^2+5x-7}-x)(\sqrt{x^2+5x-7}+x)}{\sqrt{x^2+5x-7}+x}$
$= \lim _{x \rightarrow \infty} \frac{x^2+5x-7-x^2}{\sqrt{x^2+5x-7}+x}$
$= \lim _{x \rightarrow \infty} \frac{5x-7}{\sqrt{x^2+5x-7}+x}$
Dividing the numerator and denominator by $x$ (where $x > 0$):
$= \lim _{x \rightarrow \infty} \frac{5-\frac{7}{x}}{\sqrt{1+\frac{5}{x}-\frac{7}{x^2}}+1}$
As $x \rightarrow \infty$,$\frac{1}{x} \rightarrow 0$ and $\frac{1}{x^2} \rightarrow 0$:
$= \frac{5-0}{\sqrt{1+0-0}+1} = \frac{5}{1+1} = \frac{5}{2}$

(C) We evaluate the limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$
First,factor the denominator: $2x^2+x-3 = (x-1)(2x+3)$
Substitute this into the limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(x-1)(2 x+3)}$
Recall that $(x-1) = (\sqrt{x}-1)(\sqrt{x}+1)$,so: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)(2 x+3)}$
Cancel the common term $(\sqrt{x}-1)$: $\lim _{x \rightarrow 1} \frac{2 x-3}{(\sqrt{x}+1)(2 x+3)}$
Now,substitute $x=1$: $\frac{2(1)-3}{(\sqrt{1}+1)(2(1)+3)} = \frac{-1}{(2)(5)} = \frac{-1}{10}$

(B) The sum of the first $n$ natural numbers is given by the formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.
Substituting this into the limit expression:
$\lim _{n \rightarrow \infty} \frac{n(n+1)}{2n^{2}}$
$= \lim _{n \rightarrow \infty} \frac{n^2+n}{2n^2}$
$= \lim _{n \rightarrow \infty} \frac{1 + \frac{1}{n}}{2}$
As $n \rightarrow \infty$,$\frac{1}{n} \rightarrow 0$.
Therefore,the limit is $\frac{1+0}{2} = \frac{1}{2}$.

(C) Given the limit: $\lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \sin k x}{x^{2}}=4$
We can rewrite the expression as: $\lim _{x \rightarrow 0} \left( \frac{e^{k x}-1}{x} \cdot \frac{\sin k x}{x} \right) = 4$
Using the standard limits $\lim _{x \rightarrow 0} \frac{e^{k x}-1}{x} = k$ and $\lim _{x \rightarrow 0} \frac{\sin k x}{x} = k$:
$k \cdot k = 4$
$k^{2} = 4$
$k = \pm 2$

(C) Let $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(8 x^3-\pi^3) \cos x}{(\pi-2 x)^4}$.
We can rewrite the expression as:
$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(2x-\pi)(4x^2+\pi^2+2\pi x) \cos x}{16(\frac{\pi}{2}-x)^4}$.
Let $x - \frac{\pi}{2} = h$,so $x = \frac{\pi}{2} + h$. As $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$.
Then $\cos x = \cos(\frac{\pi}{2} + h) = -\sin h$ and $1 - \sin x = 1 - \sin(\frac{\pi}{2} + h) = 1 - \cos h$.
Also,$2x - \pi = 2(\frac{\pi}{2} + h) - \pi = 2h$.
Substituting these into the limit:
$L = \lim _{h}$ ${\rightarrow 0} \frac{(1-\cos h)(2h)(4(\frac{\pi}{2}+h)^2 + \pi^2 + 2\pi(\frac{\pi}{2}+h))(-\sin h)}{16(-h)^4}$.
$L = \lim _{h \rightarrow 0} \frac{(1-\cos h)(2h)(3\pi^2 + 8\pi h + 4h^2)(-\sin h)}{16h^4}$.
$L = \lim _{h \rightarrow 0} \frac{-(1-\cos h)}{h^2} \cdot \frac{\sin h}{h} \cdot \frac{2h(3\pi^2 + 8\pi h + 4h^2)}{16h}$.
$L = -(\frac{1}{2}) \cdot (1) \cdot \frac{2(3\pi^2)}{16} = -\frac{1}{2} \cdot \frac{6\pi^2}{16} = -\frac{3\pi^2}{16}$.

(C) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$.
Using the identity $1 - \cos 2x = 2 \sin^2 x$,the denominator becomes $(2 \sin^2 x)^2 = 4 \sin^4 x$.
So,the expression is $\lim _{x \rightarrow 0} \frac{x(\tan 2x - 2 \tan x)}{4 \sin^4 x}$.
Using Taylor series expansions: $\tan \theta = \theta + \frac{\theta^3}{3} + \dots$ and $\sin x \approx x$.
$\tan 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$.
$2 \tan x = 2(x + \frac{x^3}{3} + O(x^5)) = 2x + \frac{2x^3}{3} + O(x^5)$.
Numerator: $x[(2x + \frac{8x^3}{3}) - (2x + \frac{2x^3}{3})] = x[\frac{6x^3}{3}] = 2x^4$.
Denominator: $4 \sin^4 x \approx 4x^4$.
Limit: $\lim _{x \rightarrow 0} \frac{2x^4}{4x^4} = \frac{2}{4} = \frac{1}{2}$.