Concept of limits, Evaluation of algebric limits Questions in English

(C) We need to evaluate the limit: $\lim _{x \rightarrow 1} \left( \lim _{y \rightarrow \infty} y \left( e^{x/y} - 1 \right) \right)$.
First,consider the inner limit: $L_{inner} = \lim _{y \rightarrow \infty} y \left( e^{x/y} - 1 \right)$.
Using the Taylor series expansion for $e^u = 1 + u + \frac{u^2}{2!} + \dots$,where $u = x/y$:
$L_{inner} = \lim _{y \rightarrow \infty} y \left( (1 + \frac{x}{y} + \frac{x^2}{2y^2} + \dots) - 1 \right)$
$L_{inner} = \lim _{y \rightarrow \infty} y \left( \frac{x}{y} + \frac{x^2}{2y^2} + \dots \right)$
$L_{inner} = \lim _{y \rightarrow \infty} \left( x + \frac{x^2}{2y} + \dots \right) = x$.
Now,evaluate the outer limit: $\lim _{x \rightarrow 1} (x) = 1$.

(A) Given,$\lim _{n \rightarrow \infty} x^n \log _e x=0$.
We know that as $n \rightarrow \infty$,$x^n \rightarrow 0$ only when $x \in (0, 1)$.
If $x > 1$,then $x^n \rightarrow \infty$,and if $x = 1$,$\log _e 1 = 0$,but the limit behavior for $x^n$ is generally considered for $x \in (0, 1)$ to ensure the product vanishes.
Thus,for the limit to be $0$,we must have $x \in (0, 1)$.
Since the base of the logarithm $\log _x 12$ is $x \in (0, 1)$ and the argument $12 > 1$,the value of $\log _x 12$ must be negative.

(D) Given the limit as $n \rightarrow \infty$:
Since $2-\sqrt{2} \approx 0.586 < 1$,we have $\lim _{n \rightarrow \infty} (2-\sqrt{2})^n = 0$.
Dividing the numerator and denominator by $(2+\sqrt{2})^n$:
$= \lim _{n}$ ${\rightarrow \infty} \sqrt{2} \left[ \frac{1 + \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right)^n}{1 - \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right)^n} \right] = \sqrt{2} \left[ \frac{1+0}{1-0} \right] = \sqrt{2}$.

(A) Case $1$: When $x < 0$,as $n \rightarrow \infty$,$nx \rightarrow -\infty$. Thus,$e^{nx} \rightarrow 0$. The limit becomes $\lim _{n \rightarrow \infty} \frac{A+0}{x+A(0)} = \frac{A}{x}$.
Case $2$: When $x > 0$,as $n \rightarrow \infty$,$nx \rightarrow \infty$. Thus,$e^{nx} \rightarrow \infty$. Dividing numerator and denominator by $e^{nx}$,we get $\lim _{n \rightarrow \infty} \frac{A/e^{nx} + 1}{x/e^{nx} + A} = \frac{0+1}{0+A} = \frac{1}{A}$ (assuming $A \neq 0$).

(A) For $k = \lim _{x \rightarrow 0^{+}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right)$,let $t = 1/x$. As $x \rightarrow 0^{+}$,$t \rightarrow \infty$.
$k = \lim _{t \rightarrow \infty} \frac{1}{t^2} \left( \frac{e^t - e^{-t}}{e^t + e^{-t}} \right) = \lim _{t \rightarrow \infty} \frac{1}{t^2} \left( \frac{1 - e^{-2t}}{1 + e^{-2t}} \right) = 0 \times 1 = 0$.
For $l = \lim _{x \rightarrow 0^{-}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right)$,let $y = 1/x$. As $x \rightarrow 0^{-}$,$y \rightarrow -\infty$.
$l = \lim _{y \rightarrow -\infty} \frac{1}{y^2} \left( \frac{e^y - e^{-y}}{e^y + e^{-y}} \right) = \lim _{y \rightarrow -\infty} \frac{1}{y^2} \left( \frac{e^{2y} - 1}{e^{2y} + 1} \right) = 0 \times (-1) = 0$.
Therefore,$k = l = 0$.

(D) To find $\lim_{x \rightarrow 0^+} f(x)$,let $x \rightarrow 0^+$,then $\frac{1}{x} \rightarrow \infty$.
Thus,$\lim_{x \rightarrow 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \lim_{x \rightarrow 0^+} \frac{1 - e^{-1/x}}{1 + e^{-1/x}} = \frac{1 - 0}{1 + 0} = 1$.
To find $\lim_{x \rightarrow 0^-} f(x)$,let $x \rightarrow 0^-$,then $\frac{1}{x} \rightarrow -\infty$.
Thus,$\lim_{x \rightarrow 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \frac{0 - 1}{0 + 1} = -1$.
Since the left-hand limit and right-hand limit are not equal,$\lim_{x \rightarrow 0} f(x)$ does not exist.
For $\lim_{x \rightarrow \infty} f(x)$,$\frac{1}{x} \rightarrow 0$,so $\lim_{x \rightarrow \infty} f(x) = \frac{e^0 - 1}{e^0 + 1} = \frac{1 - 1}{1 + 1} = 0$.
Therefore,option $D$ is correct.

(C) Given the limit: $\lim _{n \rightarrow \infty} \frac{1-10^n}{1+10^{n+1}}=-\frac{\alpha}{10}$
Divide the numerator and denominator by $10^n$:
$\lim _{n \rightarrow \infty} \frac{\frac{1}{10^n}-1}{\frac{1}{10^n}+10}=-\frac{\alpha}{10}$
As $n \rightarrow \infty$,$\frac{1}{10^n} \rightarrow 0$.
Substituting this value,we get:
$\frac{0-1}{0+10} = -\frac{\alpha}{10}$
$-\frac{1}{10} = -\frac{\alpha}{10}$
Therefore,$\alpha = 1$.

(A) Given,$\lim _{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{q x}=e^9$.
Using the standard limit formula $\lim _{x \rightarrow \infty}(1+\frac{a}{x})^x = e^a$,we have:
$\lim _{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{q x} = \left[ \lim _{x \rightarrow \infty} \left(1+\frac{p}{x}\right)^{x} \right]^{q} = (e^p)^q = e^{pq}$.
Comparing this with the given expression $e^9$,we get $pq = 9$.
Since $p, q \in \mathbb{N}$ (natural numbers),the possible pairs $(p, q)$ are $(1, 9), (3, 3), (9, 1)$.
In these cases,$p+q$ can be $1+9=10$ or $3+3=6$.
Given the options,the correct value is $6$.

(B) We know that the standard limit formula is $\lim _{x \rightarrow 0}(1+ax)^{\frac{1}{ax}} = e$ for $a \neq 0$.
Given expression is $\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}}$.
We can rewrite the exponent as $\frac{2}{x} = 6 \times \frac{1}{3x}$.
Substituting this into the limit,we get $\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = \lim _{x \rightarrow 0} \left((1+3x)^{\frac{1}{3x}}\right)^6$.
Using the standard limit property,this expression evaluates to $e^6$.

(B) Given the limit: $\lim _{n \rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}}$.
Since $x < 0$,as $n \rightarrow \infty$,$e^{n x} \rightarrow 0$.
Dividing the numerator and denominator by $e^{n x}$ (or simply evaluating the limit as $e^{n x} \rightarrow 0$):
$\lim _{n}$ ${\rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}} = \frac{\sin x - 0}{1 + A(0)} = \frac{\sin x}{1} = \sin x$.
Wait,let us re-evaluate the limit expression: $\lim _{n \rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}}$.
If $x < 0$,then $e^{n x} \rightarrow 0$ as $n \rightarrow \infty$.
Thus,$\frac{\sin x - 0}{1 + A(0)} = \sin x$.

(B) Let $P = \lim _{n}$ ${\rightarrow \infty} n^{-n k} \left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-1}}\right)\right\}^n$.
Taking the natural logarithm on both sides:
$\log P = \lim _{n \rightarrow \infty} n \left[ \sum_{j=0}^{k-1} \log \left(1 + \frac{1}{2^j n}\right) \right]$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\log(1+ax)}{x} = a$,we have:
$\log P = \sum_{j=0}^{k-1} \lim _{n}$ ${\rightarrow \infty} \frac{\log \left(1 + \frac{1}{2^j n}\right)}{1/n} = \sum_{j=0}^{k-1} \frac{1}{2^j}$.
This is a geometric series with $k$ terms,first term $a=1$ and common ratio $r=1/2$:
$\sum_{j=0}^{k-1} \left(\frac{1}{2}\right)^j = \frac{1(1-(1/2)^k)}{1-1/2} = 2 \left(1 - \frac{1}{2^k}\right)$.
Therefore,$P = e^{2 \left(1 - \frac{1}{2^k}\right)}$.

(A) Let $l = \lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}$.
We can factor the numerator as follows:
$6^x - 3^x - 2^x + 1 = 3^x(2^x - 1) - 1(2^x - 1) = (2^x - 1)(3^x - 1)$.
Substituting this back into the limit:
$l = \lim _{x \rightarrow 0} \frac{(2^x - 1)(3^x - 1)}{x^2} = \lim _{x \rightarrow 0} \left( \frac{2^x - 1}{x} \right) \times \left( \frac{3^x - 1}{x} \right)$.
Using the standard limit formula $\lim _{x \rightarrow 0} \frac{a^x - 1}{x} = \log _e a$,we get:
$l = (\log _e 2) \times (\log _e 3)$.

(C) The given limit is of the form $1^\infty$.
We use the formula $\lim _{x \rightarrow \infty} (f(x))^{g(x)} = e^{\lim _{x \rightarrow \infty} (f(x)-1)g(x)}$.
Here,$f(x) = \frac{3x^2-2x+3}{3x^2+x-2}$ and $g(x) = 3x-2$.
$(f(x)-1) = \frac{3x^2-2x+3 - (3x^2+x-2)}{3x^2+x-2} = \frac{-3x+5}{3x^2+x-2}$.
Now,$\lim _{x \rightarrow \infty} (f(x)-1)g(x) = \lim _{x \rightarrow \infty} \left(\frac{-3x+5}{3x^2+x-2}\right)(3x-2)$.
$= \lim _{x \rightarrow \infty} \frac{-9x^2+6x+15x-10}{3x^2+x-2} = \lim _{x \rightarrow \infty} \frac{-9x^2+21x-10}{3x^2+x-2}$.
Dividing the numerator and denominator by $x^2$,we get $\frac{-9}{3} = -3$.
Therefore,the limit is $e^{-3}$.

(C) Let $L = \lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n$.
Taking the natural logarithm on both sides:
$\ln L = \lim _{n \rightarrow \infty} n \ln \left(1 + \frac{b^{1 / n} - 1}{a}\right)$.
Using the limit formula $\lim _{x \rightarrow 0} \frac{\ln(1 + x)}{x} = 1$,let $x = \frac{b^{1 / n} - 1}{a}$. As $n \rightarrow \infty$,$x \rightarrow 0$.
$\ln L = \lim _{n \rightarrow \infty} n \cdot \left(\frac{b^{1 / n} - 1}{a}\right) \cdot \frac{\ln(1 + x)}{x} = \lim _{n \rightarrow \infty} \frac{n}{a} (b^{1 / n} - 1)$.
Let $t = 1/n$. As $n \rightarrow \infty$,$t \rightarrow 0$.
$\ln L = \frac{1}{a} \lim _{t \rightarrow 0} \frac{b^t - 1}{t}$.
Since $\lim _{t \rightarrow 0} \frac{b^t - 1}{t} = \ln b$,we have:
$\ln L = \frac{1}{a} \ln b = \ln(b^{1 / a})$.
Therefore,$L = b^{1 / a}$.

(D) Given,$f(x) = \lim_{y \rightarrow \infty} y(x^{1/y} - 1)$.
First,we simplify $f(x)$:
$f(x) = \lim_{y \rightarrow \infty} \frac{x^{1/y} - 1}{1/y}$.
Let $t = 1/y$. As $y \rightarrow \infty$,$t \rightarrow 0^+$.
$f(x) = \lim_{t \rightarrow 0} \frac{x^t - 1}{t}$.
Using the standard limit $\lim_{t \rightarrow 0} \frac{a^t - 1}{t} = \ln(a)$,we get $f(x) = \ln(x)$.
Now,substitute $f(x) = \ln(x)$ into the given equation $2022 f(\frac{1}{x}) + P f(x) = f(x^2)$:
$2022 \ln(\frac{1}{x}) + P \ln(x) = \ln(x^2)$.
Since $\ln(\frac{1}{x}) = -\ln(x)$ and $\ln(x^2) = 2 \ln(x)$:
$2022(-\ln(x)) + P \ln(x) = 2 \ln(x)$.
$-2022 \ln(x) + P \ln(x) = 2 \ln(x)$.
Dividing by $\ln(x)$ (for $x \neq 1$):
$-2022 + P = 2$.
$P = 2024$.

(A) We use the standard limit formula: $\lim _{x \rightarrow \infty} (1 + \frac{a}{x})^{bx} = e^{ab}$.
Given expression: $\lim _{x \rightarrow \infty} (1 + \frac{2}{x})^{3x}$.
Here,$a = 2$ and $b = 3$.
Applying the formula: $e^{2 \times 3} = e^6$.

(A) Let $L = \lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}$.
Rationalizing the numerator,we multiply by $\frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$:
$L = \lim _{y \rightarrow 0} \frac{(1+\sqrt{1+y^4})-2}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})} = \lim _{y \rightarrow 0} \frac{\sqrt{1+y^4}-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})}$.
Rationalizing the numerator again,we multiply by $\frac{\sqrt{1+y^4}+1}{\sqrt{1+y^4}+1}$:
$L = \lim _{y \rightarrow 0} \frac{(1+y^4)-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)} = \lim _{y \rightarrow 0} \frac{y^4}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$.
Canceling $y^4$ and evaluating the limit as $y \rightarrow 0$:
$L = \frac{1}{(\sqrt{1+1}+\sqrt{2})(\sqrt{1}+1)} = \frac{1}{(\sqrt{2}+\sqrt{2})(1+1)} = \frac{1}{2\sqrt{2} \times 2} = \frac{1}{4\sqrt{2}}$.

(D) Let $x = 3 - h$,where $h \rightarrow 0^{+}$.
As $x \rightarrow 3^{-}$,$|x| = x = 3 - h$ and $|3-x| = |3-(3-h)| = |h| = h$.
Also,$[3x-9] = [3(3-h)-9] = [9-3h-9] = [-3h]$. Since $h > 0$,$-3h$ is a small negative number,so $[-3h] = -1$.
And $[9-3x] = [9-3(3-h)] = [9-9+3h] = [3h]$. Since $h > 0$,$3h$ is a small positive number,so $[3h] = 0$.
Substituting these into the limit expression:
$\lim _{h \rightarrow 0^{+}} \frac{(3-(3-h)+\sin h) \cos(0)}{h(-1)} = \lim _{h \rightarrow 0^{+}} \frac{(h+\sin h)(1)}{-h} = \lim _{h \rightarrow 0^{+}} -\left(1 + \frac{\sin h}{h}\right) = -(1+1) = -2$.

(D) Let $f(x) = \frac{|2 \cos x - 1|}{2 \cos x - 1}$ for $0 \leq x \leq \pi/2$.
We know that $2 \cos x - 1 > 0$ when $\cos x > 1/2$,i.e.,$0 \leq x < \pi/3$,and $2 \cos x - 1 < 0$ when $\pi/3 < x \leq \pi/2$.
Thus,$f(x) = 1$ for $0 \leq x < \pi/3$ and $f(x) = -1$ for $\pi/3 < x \leq \pi/2$.
For the limit to exist at $x=a$,the left-hand limit and right-hand limit must be equal.
For $0 \leq a < \pi/3$,the function is constant $1$,so the limit is $1$.
For $\pi/3 < a \leq \pi/2$,the function is constant $-1$,so the limit is $-1$.
At $a = \pi/3$,the left-hand limit is $1$ and the right-hand limit is $-1$,so the limit does not exist.
Therefore,the correct statement is that the limit is $1$ when $0 \leq a < \pi/3$.

(A) $\text{Given limit: } L = \lim _{x \rightarrow \frac{3}{2}} \frac{2x(2x-3)(4x^2+6x+9)}{(2x)^{1/3} - 3^{1/3}}$
$\text{Rationalizing the denominator using } a^3 - b^3 = (a-b)(a^2+ab+b^2) \text{ where } a=(2x)^{1/3}, b=3^{1/3}:$
$L = \lim _{x \rightarrow \frac{3}{2}} \frac{2x(2x-3)(4x^2+6x+9)((2x)^{2/3} + (6x)^{1/3} + 3^{2/3})}{2x-3}$
$L = \lim _{x \rightarrow \frac{3}{2}} 2x(4x^2+6x+9)((2x)^{2/3} + (6x)^{1/3} + 3^{2/3})$
$\text{Substituting } x = \frac{3}{2}:$
$L = 2(\frac{3}{2})(4(\frac{9}{4}) + 6(\frac{3}{2}) + 9)(3^{2/3} + 9^{1/3} + 3^{2/3})$
$L = 3(9 + 9 + 9)(3 \times 3^{2/3})$
$L = 3(27)(3^{5/3}) = 3^1 \times 3^3 \times 3^{5/3} = 3^{1+3+5/3} = 3^{17/3} = \sqrt[3]{3^{17}}$

(B) Let $L = \lim _{x \rightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{1 - \cos 4x}$.
Using the identity $1 - \cos 4x = 2 \sin^2 2x$,we have $L = \lim _{x \rightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{2 \sin^2 2x}$.
Divide numerator and denominator by $x^2$ and use standard limits $\lim_{x \to 0} \frac{\sin ax}{ax} = 1$:
$L = \lim _{x \rightarrow 0} \left( \frac{3^{2x} - \sqrt{x+1}}{x} \right) \cdot \left( \frac{\sin 5x}{x} \right) \cdot \left( \frac{x^2}{2 \sin^2 2x} \right) = \lim _{x \rightarrow 0} \left( \frac{3^{2x} - 1 - (\sqrt{x+1} - 1)}{x} \right) \cdot 5 \cdot \frac{1}{2(2)^2}$.
$L = \frac{5}{8} \lim _{x \rightarrow 0} \left( \frac{3^{2x} - 1}{x} - \frac{\sqrt{x+1} - 1}{x} \right)$.
Using $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$ and $\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} = \frac{1}{2}$:
$L = \frac{5}{8} \left( 2 \ln 3 - \frac{1}{2} \right) = \frac{5}{16} (4 \ln 3 - 1) = \frac{5}{16} (\ln 81 - \ln e) = \frac{5}{16} \ln \left( \frac{81}{e} \right)$.

(A) For the first limit: $\lim_{x \rightarrow -2^{+}} [x] = -2$.
Thus,$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) = (-2)^2 - (-2) - 2 = 4 + 2 - 2 = 4$.
For the second limit: $\lim_{x \rightarrow -3^{-}} [x] = -4$.
Thus,$\lim_{x \rightarrow -3^{-}} ([x]^2 - 4[x] + 3) = (-4)^2 - 4(-4) + 3 = 16 + 16 + 3 = 35$.
Adding both results: $4 + 35 = 39$.

(D) Let $L = \lim _{x \rightarrow 0} \frac{\tan 2x - 2\tan x}{(1 - \cos x)(2^x - 1)}$.
Using the expansion $\tan x = x + \frac{x^3}{3} + O(x^5)$:
$\tan 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$.
$2\tan x = 2(x + \frac{x^3}{3} + O(x^5)) = 2x + \frac{2x^3}{3} + O(x^5)$.
Numerator: $\tan 2x - 2\tan x = (2x + \frac{8x^3}{3}) - (2x + \frac{2x^3}{3}) = \frac{6x^3}{3} = 2x^3$.
Denominator: $(1 - \cos x)(2^x - 1) \approx (\frac{x^2}{2})(x \ln 2) = \frac{x^3 \ln 2}{2}$.
$L = \lim _{x \rightarrow 0} \frac{2x^3}{\frac{x^3 \ln 2}{2}} = \frac{2 \times 2}{\ln 2} = \frac{4}{\ln 2}$.

(A) Given $f(x) = 2x - [2x]$.
To find $l_1 = \lim_{x \rightarrow 2^{-}} f(x)$:
Let $x = 2 - h$,where $h \rightarrow 0$ and $h > 0$.
$l_1 = \lim_{h \rightarrow 0} (2(2 - h) - [2(2 - h)]) = \lim_{h \rightarrow 0} (4 - 2h - [4 - 2h])$.
Since $h$ is a very small positive number,$4 - 2h$ is slightly less than $4$,so $[4 - 2h] = 3$.
Thus,$l_1 = 4 - 3 = 1$.
To find $l_2 = \lim_{x \rightarrow 2^{+}} f(x)$:
Let $x = 2 + h$,where $h \rightarrow 0$ and $h > 0$.
$l_2 = \lim_{h \rightarrow 0} (2(2 + h) - [2(2 + h)]) = \lim_{h \rightarrow 0} (4 + 2h - [4 + 2h])$.
Since $h$ is a very small positive number,$4 + 2h$ is slightly greater than $4$,so $[4 + 2h] = 4$.
Thus,$l_2 = 4 - 4 = 0$.
Therefore,$l_1 + l_2 = 1 + 0 = 1$.

(B) We are given the function $f(x) = \frac{\sin(1+[x])}{[x]}$ for $[x] \neq 0$.
To find $\lim_{x \rightarrow 0^{-}} f(x)$,we consider the values of $x$ slightly less than $0$.
For $x \in (-1, 0)$,the greatest integer function $[x] = -1$.
Substituting this into the limit:
$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} \frac{\sin(1+[x])}{[x]} = \frac{\sin(1+(-1))}{-1} = \frac{\sin(0)}{-1} = \frac{0}{-1} = 0$.

(D) First,evaluate $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x(a^x - 1)}{1 - \cos x}$.
Using the standard limits $\lim_{x \to 0} \frac{a^x - 1}{x} = \log a$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$,we rewrite $f(x)$ as $\frac{(a^x - 1)/x}{(1 - \cos x)/x^2} = \frac{\log a}{1/2} = 2 \log a$.
Next,evaluate $\lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{x(1 - a^x)}{a^x(\sqrt{1 - x^2} - \sqrt{1 + x^2})}$.
Rationalize the denominator: $\sqrt{1 - x^2} - \sqrt{1 + x^2} = \frac{(1 - x^2) - (1 + x^2)}{\sqrt{1 - x^2} + \sqrt{1 + x^2}} = \frac{-2x^2}{\sqrt{1 - x^2} + \sqrt{1 + x^2}}$.
So,$g(x) = \frac{x(1 - a^x)}{a^x \cdot \frac{-2x^2}{\sqrt{1 - x^2} + \sqrt{1 + x^2}}} = \frac{-(a^x - 1)}{x} \cdot \frac{\sqrt{1 - x^2} + \sqrt{1 + x^2}}{-2 a^x} = \frac{a^x - 1}{x} \cdot \frac{\sqrt{1 - x^2} + \sqrt{1 + x^2}}{2 a^x}$.
Taking the limit as $x \to 0$: $\lim_{x \to 0} g(x) = (\log a) \cdot \frac{1 + 1}{2(1)} = \log a$.
Finally,$\lim_{x \to 0} (f(x) - g(x)) = 2 \log a - \log a = \log a$.

(B) Given the limit $L = \lim _{x \rightarrow 0} \frac{3^{x^3}-\left(1-x^3\right)^{2 / 3}}{x^2 \sin x}$.
Since $\sin x \approx x$ as $x \rightarrow 0$, the expression becomes $\lim _{x \rightarrow 0} \frac{3^{x^3}-\left(1-x^3\right)^{2 / 3}}{x^3}$.
Using the series expansions $a^u = 1 + u \ln a + O(u^2)$ and $(1+u)^n = 1 + nu + O(u^2)$, we have:
$3^{x^3} = 1 + x^3 \ln 3 + O(x^6)$
$(1-x^3)^{2/3} = 1 - \frac{2}{3}x^3 + O(x^6)$
Substituting these into the limit:
$L = \lim _{x \rightarrow 0} \frac{(1 + x^3 \ln 3) - (1 - \frac{2}{3}x^3)}{x^3} = \lim _{x \rightarrow 0} \frac{x^3(\ln 3 + \frac{2}{3})}{x^3} = \ln 3 + \frac{2}{3} = \frac{2}{3} + \log_e 3$.
Comparing this with $p + \log q$, we get $p = \frac{2}{3}$ and $q = 3$.
Therefore, $pq = \frac{2}{3} \times 3 = 2$.

(B) Let $L = \lim _{x \rightarrow 0} \frac{(\cos x)^{1/2} - (\cos x)^{1/3}}{\sin ^2 x}$.
Using the Taylor series expansion for $\cos x \approx 1 - \frac{x^2}{2}$ and $\sin x \approx x$ as $x \rightarrow 0$:
$L = \lim _{x \rightarrow 0} \frac{(1 - \frac{x^2}{2})^{1/2} - (1 - \frac{x^2}{2})^{1/3}}{x^2}$.
Using the binomial expansion $(1+u)^n \approx 1 + nu$ for small $u$:
$(1 - \frac{x^2}{2})^{1/2} \approx 1 - \frac{1}{2}(\frac{x^2}{2}) = 1 - \frac{x^2}{4}$.
$(1 - \frac{x^2}{2})^{1/3} \approx 1 - \frac{1}{3}(\frac{x^2}{2}) = 1 - \frac{x^2}{6}$.
Substituting these into the limit:
$L = \lim _{x \rightarrow 0} \frac{(1 - \frac{x^2}{4}) - (1 - \frac{x^2}{6})}{x^2} = \lim _{x \rightarrow 0} \frac{-\frac{x^2}{4} + \frac{x^2}{6}}{x^2} = -\frac{1}{4} + \frac{1}{6} = \frac{-3 + 2}{12} = -\frac{1}{12}$.

(B) Let $x = \tan \theta$. As $\theta \rightarrow \frac{\pi}{2}^{-}$,$x \rightarrow \infty$.
The expression becomes $\lim _{x \rightarrow \infty} \frac{8x^4 + 4x^2 + 5}{(3-2x)^4}$.
Dividing the numerator and denominator by $x^4$:
$\lim _{x \rightarrow \infty} \frac{8 + \frac{4}{x^2} + \frac{5}{x^4}}{(\frac{3}{x} - 2)^4}$.
As $x \rightarrow \infty$,$\frac{4}{x^2} \rightarrow 0$,$\frac{5}{x^4} \rightarrow 0$,and $\frac{3}{x} \rightarrow 0$.
Thus,the limit is $\frac{8 + 0 + 0}{(0 - 2)^4} = \frac{8}{(-2)^4} = \frac{8}{16} = \frac{1}{2}$.

(B) Given limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$
Factorize the denominator: $2x^2+x-3 = (2x+3)(x-1)$
Substitute this into the limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(2 x+3)(x-1)}$
Multiply the numerator and denominator by $(\sqrt{x}+1)$: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)(\sqrt{x}+1)}{(2 x+3)(x-1)(\sqrt{x}+1)}$
Simplify using $(\sqrt{x}-1)(\sqrt{x}+1) = (x-1)$: $\lim _{x \rightarrow 1} \frac{(2 x-3)(x-1)}{(2 x+3)(x-1)(\sqrt{x}+1)}$
Cancel $(x-1)$ from numerator and denominator: $\lim _{x \rightarrow 1} \frac{2 x-3}{(2 x+3)(\sqrt{x}+1)}$
Evaluate the limit by substituting $x=1$: $\frac{2(1)-3}{(2(1)+3)(\sqrt{1}+1)} = \frac{-1}{5(2)} = -\frac{1}{10}$

(A) Let $L = \lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2}$.
Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$,where $a = \sqrt[3]{6+x}$ and $b = \sqrt[3]{10-x}$:
$L = \lim _{x \rightarrow 2} \frac{(6+x) - (10-x)}{(x-2)((6+x)^{2/3} + (6+x)^{1/3}(10-x)^{1/3} + (10-x)^{2/3})}$
$L = \lim _{x \rightarrow 2} \frac{2x - 4}{(x-2)((6+x)^{2/3} + (6+x)^{1/3}(10-x)^{1/3} + (10-x)^{2/3})}$
$L = \lim _{x \rightarrow 2} \frac{2(x-2)}{(x-2)((6+x)^{2/3} + (6+x)^{1/3}(10-x)^{1/3} + (10-x)^{2/3})}$
$L = \frac{2}{8^{2/3} + (8 \times 8)^{1/3} + 8^{2/3}} = \frac{2}{4 + 4 + 4} = \frac{2}{12} = \frac{1}{6}$.

(B) Given,$\lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2}$
$= \lim _{x \rightarrow 0} \frac{(2^x)^2 - 2 \cdot 2^x + 1 + 1 - \cos 2x}{x^2}$
$= \lim _{x \rightarrow 0} \frac{(2^x - 1)^2 + (1 - \cos 2x)}{x^2}$
$= \lim _{x \rightarrow 0} \left( \frac{2^x - 1}{x} \right)^2 + \lim _{x \rightarrow 0} \frac{1 - \cos 2x}{x^2}$
Using the standard limits $\lim _{x \rightarrow 0} \frac{a^x - 1}{x} = \log _e a$ and $\lim _{x \rightarrow 0} \frac{1 - \cos kx}{x^2} = \frac{k^2}{2}$:
$= (\log _e 2)^2 + \frac{2^2}{2}$
$= (\log _e 2)^2 + 2$

(C) Let $L = \lim _{x \rightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right]$.
Rationalizing the expression:
$L = \lim _{x \rightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right] \times \frac{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x}{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x}$
$= \lim _{x \rightarrow \infty} x^3 \frac{x^2 + \sqrt{x^4 + 1} - 2x^2}{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x} = \lim _{x \rightarrow \infty} x^3 \frac{\sqrt{x^4 + 1} - x^2}{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x}$.
Rationalizing again:
$= \lim _{x \rightarrow \infty} x^3 \frac{(\sqrt{x^4 + 1} - x^2)(\sqrt{x^4 + 1} + x^2)}{(\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x)(\sqrt{x^4 + 1} + x^2)} = \lim _{x \rightarrow \infty} x^3 \frac{x^4 + 1 - x^4}{(\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x)(\sqrt{x^4 + 1} + x^2)}$
$= \lim _{x \rightarrow \infty} \frac{x^3}{(\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x)(\sqrt{x^4 + 1} + x^2)}$.
Dividing numerator and denominator by $x^3$ (specifically $x$ in the first bracket and $x^2$ in the second):
$= \lim _{x \rightarrow \infty} \frac{1}{(\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2})(\sqrt{1 + \frac{1}{x^4}} + 1)} = \frac{1}{(\sqrt{1 + 1} + \sqrt{2})(1 + 1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$.

(C) Given that $f(x)$ is a differentiable function with $f(0)=0$ and $f^{\prime}(0)=20$.
We need to find $\lim _{x \rightarrow 0} A(x) = \lim _{x \rightarrow 0} [2 f(x) \operatorname{cosec} 4 x + 4 f(x)(\cos ^2 x + 1) - 4 \cos ^2 x]$.
Rewrite the expression as:
$\lim _{x \rightarrow 0} A(x) = \lim _{x \rightarrow 0} \left[ \frac{2 f(x)}{\sin 4x} + 4 f(x)(\cos ^2 x + 1) - 4 \cos ^2 x \right]$.
Since $f(0)=0$,the first term is an indeterminate form of type $\frac{0}{0}$.
Using the limit $\lim _{x \rightarrow 0} \frac{f(x)}{\sin 4x} = \lim _{x \rightarrow 0} \frac{f(x)}{x} \cdot \frac{4x}{\sin 4x} \cdot \frac{1}{4} = f^{\prime}(0) \cdot 1 \cdot \frac{1}{4} = \frac{20}{4} = 5$.
Now,evaluate the limit:
$\lim _{x \rightarrow 0} A(x) = 2 \left( \lim _{x \rightarrow 0} \frac{f(x)}{\sin 4x} \right) + 4 \lim _{x \rightarrow 0} f(x)(\cos ^2 x + 1) - 4 \lim _{x \rightarrow 0} \cos ^2 x$.
$= 2(5) + 4(0)(1+1) - 4(1)^2$.
$= 10 + 0 - 4 = 6$.
Therefore,the correct option is $C$.

(D) We need to evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}$.
First,consider the term $(1+\sin x)^{\frac{2}{\sin x}}$. As $x \rightarrow 0$,$\sin x \rightarrow 0$,so this is of the form $(1+u)^{2/u}$ where $u = \sin x$. We know that $\lim _{u \rightarrow 0} (1+u)^{1/u} = e$,so $\lim _{x \rightarrow 0} (1+\sin x)^{\frac{2}{\sin x}} = e^2$.
Now,rewrite the limit as:
$L = \lim _{x \rightarrow 0} \left( \frac{2^x-1}{\log(1+2x)} \right) \times \lim _{x \rightarrow 0} (1+\sin x)^{\frac{2}{\sin x}}$.
Using the standard limits $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \log a$ and $\lim _{x \rightarrow 0} \frac{\log(1+x)}{x} = 1$:
$\lim _{x \rightarrow 0} \frac{2^x-1}{\log(1+2x)} = \lim _{x \rightarrow 0} \left( \frac{2^x-1}{x} \cdot \frac{2x}{\log(1+2x)} \cdot \frac{1}{2} \right) = \log 2 \cdot 1 \cdot \frac{1}{2} = \frac{\log 2}{2} = \log \sqrt{2}$.
Thus,$L = \log \sqrt{2} \cdot e^2 = e^2 \log \sqrt{2}$.
Therefore,option $(D)$ is correct.

(D) Given the limit: $\lim _{x \rightarrow-\infty} \frac{3|x|^3-x^2+2|x|-5}{-5|x|^3+3 x^2-2|x|+7}$
Since $x \rightarrow -\infty$,we have $|x| = -x$. Let $x = -t$,where $t \rightarrow \infty$. Then $|x| = t$.
Substituting these into the expression:
$= \lim _{t \rightarrow \infty} \frac{3t^3 - (-t)^2 + 2t - 5}{-5t^3 + 3(-t)^2 - 2t + 7}$
$= \lim _{t \rightarrow \infty} \frac{3t^3 - t^2 + 2t - 5}{-5t^3 + 3t^2 - 2t + 7}$
Divide the numerator and denominator by $t^3$:
$= \lim _{t}$ ${\rightarrow \infty} \frac{3 - \frac{1}{t} + \frac{2}{t^2} - \frac{5}{t^3}}{-5 + \frac{3}{t} - \frac{2}{t^2} + \frac{7}{t^3}}$
As $t \rightarrow \infty$,all terms with $t$ in the denominator approach $0$.
$= \frac{3 - 0 + 0 - 0}{-5 + 0 - 0 + 0} = -\frac{3}{5}$

(C) We have,$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi x+2\pi\right)}{x^2}$
Since $\cos(2\pi + \theta) = \cos \theta$,the expression becomes $\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi x\right)}{x^2}$
Using the identity $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$,we get:
$\lim _{x \rightarrow 0} \frac{2 \sin^2 \left(\frac{x^2+\pi x}{2}\right)}{x^2}$
$= 2 \lim _{x}$ ${\rightarrow 0} \left[ \frac{\sin \left(\frac{x^2+\pi x}{2}\right)}{\frac{x^2+\pi x}{2}} \right]^2 \times \left( \frac{x^2+\pi x}{2} \right)^2 \times \frac{1}{x^2}$
$= 2 \times 1^2 \times \lim _{x \rightarrow 0} \frac{x^2(x+\pi)^2}{4x^2}$
$= 2 \times \frac{1}{4} \times \pi^2 = \frac{\pi^2}{2}$

(C) Given that $[x]$ is the greatest integer function.
We need to evaluate the limit: $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)$.
Let $x = 2 + h$,where $h \rightarrow 0^{+}$.
As $x \rightarrow 2^{+}$,we have $2 < x < 3$,which implies $[x] = 2$.
Also,for $2 < x < 3$,we have $\frac{2}{3} < \frac{x}{3} < 1$.
Since $\frac{2}{3} < \frac{x}{3} < 1$,the greatest integer value $\left[\frac{x}{3}\right] = 0$.
Substituting these values into the limit:
$L = \lim _{h}$ ${\rightarrow 0^{+}}\left(\frac{[2+h]^3}{3}-\left[\frac{2+h}{3}\right]^3\right) = \frac{2^3}{3} - 0^3 = \frac{8}{3}$.

(C) To evaluate the limit $\lim _{x \rightarrow 0} \frac{\sqrt{x^2+100}-10}{x^2}$,we rationalize the numerator:
$\lim _{x \rightarrow 0} \frac{\sqrt{x^2+100}-10}{x^2} \times \frac{\sqrt{x^2+100}+10}{\sqrt{x^2+100}+10}$
$= \lim _{x \rightarrow 0} \frac{(x^2+100)-100}{x^2(\sqrt{x^2+100}+10)}$
$= \lim _{x \rightarrow 0} \frac{x^2}{x^2(\sqrt{x^2+100}+10)}$
$= \lim _{x \rightarrow 0} \frac{1}{\sqrt{x^2+100}+10}$
Substituting $x = 0$:
$= \frac{1}{\sqrt{0+100}+10} = \frac{1}{10+10} = \frac{1}{20} = 0.05$
Thus,the correct option is $C$.

(D) To evaluate the limit $\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}}$,we rationalize the numerator and the denominator:
$\lim _{x}$ ${\rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 a}}{\sqrt{x}-\sqrt{a}} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}} \times \frac{\sqrt{a+2 x}+\sqrt{3 a}}{\sqrt{a+2 x}+\sqrt{3 a}}$
$= \lim _{x \rightarrow a} \frac{(a+2x) - 3a}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{a+2x}+\sqrt{3a}}$
$= \lim _{x \rightarrow a} \frac{2x - 2a}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{a+2x}+\sqrt{3a}}$
$= \lim _{x \rightarrow a} \frac{2(x-a)}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{a+2x}+\sqrt{3a}}$
$= 2 \times \frac{\sqrt{a}+\sqrt{a}}{\sqrt{a+2a}+\sqrt{3a}} = 2 \times \frac{2\sqrt{a}}{\sqrt{3a}+\sqrt{3a}} = 2 \times \frac{2\sqrt{a}}{2\sqrt{3a}} = \frac{2}{\sqrt{3}}$

(B) Given the limit: $\lim _{x \rightarrow 0} \frac{\cos 4 x-4 \cos 2 x+3}{x^4}$ which is in the $\frac{0}{0}$ indeterminate form.
Using Taylor series expansion for $\cos \theta \approx 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!}$:
$\cos 4x \approx 1 - \frac{(4x)^2}{2} + \frac{(4x)^4}{24} = 1 - 8x^2 + \frac{256x^4}{24} = 1 - 8x^2 + \frac{32x^4}{3}$.
$\cos 2x \approx 1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} = 1 - 2x^2 + \frac{16x^4}{24} = 1 - 2x^2 + \frac{2x^4}{3}$.
Substituting these into the numerator:
$(1 - 8x^2 + \frac{32x^4}{3}) - 4(1 - 2x^2 + \frac{2x^4}{3}) + 3$
$= 1 - 8x^2 + \frac{32x^4}{3} - 4 + 8x^2 - \frac{8x^4}{3} + 3$
$= (1 - 4 + 3) + (-8x^2 + 8x^2) + (\frac{32x^4}{3} - \frac{8x^4}{3})$
$= 0 + 0 + \frac{24x^4}{3} = 8x^4$.
Thus,$\lim _{x \rightarrow 0} \frac{8x^4}{x^4} = 8$.

(A) We have,$\lim _{y \rightarrow 1}\left(\frac{1}{y^2-1}-\frac{2}{y^4-1}\right)$
$= \lim _{y \rightarrow 1}\left(\frac{y^2+1-2}{y^4-1}\right)$
$= \lim _{y \rightarrow 1}\left(\frac{y^2-1}{(y^2-1)(y^2+1)}\right)$
$= \lim _{y \rightarrow 1} \frac{1}{y^2+1}$
$= \frac{1}{1^2+1} = \frac{1}{2}$

(C) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$.
Rationalizing the numerator,we multiply by $\frac{\sqrt{1+x^2}+\sqrt{1-x+x^2}}{\sqrt{1+x^2}+\sqrt{1-x+x^2}}$:
$= \lim _{x \rightarrow 0} \frac{(1+x^2)-(1-x+x^2)}{(3^x-1)(\sqrt{1+x^2}+\sqrt{1-x+x^2})}$
$= \lim _{x \rightarrow 0} \frac{x}{(3^x-1)(\sqrt{1+x^2}+\sqrt{1-x+x^2})}$
$= \lim _{x}$ ${\rightarrow 0} \left( \frac{1}{\frac{3^x-1}{x}} \right) \times \left( \frac{1}{\sqrt{1+x^2}+\sqrt{1-x+x^2}} \right)$
Using the standard limit $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \log _e a$,we get:
$= \frac{1}{\log _e 3} \times \frac{1}{\sqrt{1+0}+\sqrt{1-0+0}}$
$= \frac{1}{\log _e 3} \times \frac{1}{1+1}$
$= \frac{1}{2 \log _e 3} = \frac{1}{\log _e 3^2} = \frac{1}{\log _e 9}$.

(C) To evaluate the limit $\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$,we rationalize the numerator:
$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8} \times \frac{\sqrt{1+\sqrt{1+x}}+2}{\sqrt{1+\sqrt{1+x}}+2}$
$= \lim _{x \rightarrow 8} \frac{1+\sqrt{1+x}-4}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)}$
$= \lim _{x \rightarrow 8} \frac{\sqrt{1+x}-3}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)}$
Now,rationalize the remaining radical in the numerator:
$= \lim _{x \rightarrow 8} \frac{\sqrt{1+x}-3}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)} \times \frac{\sqrt{1+x}+3}{\sqrt{1+x}+3}$
$= \lim _{x \rightarrow 8} \frac{1+x-9}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$
$= \lim _{x \rightarrow 8} \frac{x-8}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$
$= \lim _{x \rightarrow 8} \frac{1}{(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$
Substitute $x = 8$:
$= \frac{1}{(\sqrt{1+\sqrt{9}}+2)(\sqrt{9}+3)}$
$= \frac{1}{(\sqrt{1+3}+2)(3+3)}$
$= \frac{1}{(2+2)(6)} = \frac{1}{4 \times 6} = \frac{1}{24}$

(D) To evaluate the limit $\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$,we rationalize the expression by multiplying and dividing by the conjugate $\sqrt{x^2+2 x-1}+x$:
$\lim _{x \rightarrow \infty}\left[\frac{(\sqrt{x^2+2 x-1}-x)(\sqrt{x^2+2 x-1}+x)}{\sqrt{x^2+2 x-1}+x}\right]$
$= \lim _{x \rightarrow \infty}\left[\frac{x^2+2 x-1-x^2}{\sqrt{x^2+2 x-1}+x}\right]$
$= \lim _{x \rightarrow \infty}\left[\frac{2x-1}{\sqrt{x^2(1+\frac{2}{x}-\frac{1}{x^2})}+x}\right]$
$= \lim _{x \rightarrow \infty}\left[\frac{x(2-\frac{1}{x})}{x(\sqrt{1+\frac{2}{x}-\frac{1}{x^2}}+1)}\right]$
$= \frac{2-0}{\sqrt{1+0-0}+1} = \frac{2}{2} = 1$

(A) First,we evaluate $\ell$:
$\ell = \lim_{\theta \rightarrow 0} \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta} = \lim_{\theta \rightarrow 0} \frac{\sin(3\theta)}{\theta} = \lim_{\theta \rightarrow 0} 3 \left( \frac{\sin(3\theta)}{3\theta} \right) = 3(1) = 3$.
Next,we evaluate $m$:
$m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} = \lim_{\theta \rightarrow 0} \frac{\tan(2\theta)}{\theta} = \lim_{\theta \rightarrow 0} 2 \left( \frac{\tan(2\theta)}{2\theta} \right) = 2(1) = 2$.
The quadratic equation with roots $\ell = 3$ and $m = 2$ is given by $x^2 - (\ell + m)x + \ell m = 0$.
$x^2 - (3 + 2)x + (3 \times 2) = 0$
$x^2 - 5x + 6 = 0$.