Finding unknown using limit Questions in English

(B) Let $A = 1 + a$.
As $a \rightarrow 0^{+}$,$A \rightarrow 1^{+}$.
The given equation is $(A^{\frac{1}{3}}-1) x^2+(A^{\frac{1}{2}}-1) x+(A^{\frac{1}{6}}-1)=0$.
Dividing the entire equation by $(A-1)$ where $A \neq 1$:
$\frac{A^{\frac{1}{3}}-1}{A-1} x^2 + \frac{A^{\frac{1}{2}}-1}{A-1} x + \frac{A^{\frac{1}{6}}-1}{A-1} = 0$.
Taking the limit as $A \rightarrow 1$ (which corresponds to $a \rightarrow 0^{+}$),we use the standard limit $\lim _{A \rightarrow 1} \frac{A^n-1}{A-1} = n$:
$\frac{1}{3} x^2 + \frac{1}{2} x + \frac{1}{6} = 0$.
Multiplying by $6$ to clear denominators:
$2 x^2 + 3 x + 1 = 0$.
Factoring the quadratic: $(2x + 1)(x + 1) = 0$.
Thus,the roots are $x = -1$ and $x = -\frac{1}{2}$.
Therefore,$\lim _{a \rightarrow 0^{+}} \alpha(a) = -1$ and $\lim _{a \rightarrow 0^{+}} \beta(a) = -\frac{1}{2}$.

(A) Given that $\lim_{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1} = \pi$.
Since the limit exists and the denominator $x^2-1$ approaches $0$ as $x \rightarrow 1$,the numerator $f(x)-2$ must also approach $0$ for the limit to be a finite value $\pi$.
Therefore,$\lim_{x \rightarrow 1} (f(x)-2) = 0$.
This implies $\lim_{x \rightarrow 1} f(x) = 2$.

(A) Given the limit $\lim _{x \rightarrow 0^{+}} \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$.
Since $x \rightarrow 0^{+}$,we have $[x] = 0$.
Substituting this into the expression:
$\lim _{x \rightarrow 0^{+}} \frac{\cos(0) - \cos(kx - 0)}{x^2} = 5$
$\lim _{x \rightarrow 0^{+}} \frac{1 - \cos(kx)}{x^2} = 5$
Using the limit formula $\lim _{\theta \rightarrow 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$,we rewrite the expression:
$\lim _{x \rightarrow 0^{+}} \frac{1 - \cos(kx)}{(kx)^2} \cdot k^2 = 5$
$\frac{1}{2} \cdot k^2 = 5$
$k^2 = 10$
$k = \sqrt{10}$.

(B) Given,$f(x) = \begin{cases} 3ax - 2b, & x > 1 \\ ax + b + 1, & x < 1 \end{cases}$
Since $\lim_{x \rightarrow 1} f(x)$ exists,the left-hand limit must equal the right-hand limit:
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x)$
Substituting the given functions:
$\lim_{x \rightarrow 1^{-}} (ax + b + 1) = \lim_{x \rightarrow 1^{+}} (3ax - 2b)$
Evaluating the limits at $x = 1$:
$a(1) + b + 1 = 3a(1) - 2b$
$a + b + 1 = 3a - 2b$
Rearranging the terms to find the relation:
$1 = 3a - a - 2b - b$
$1 = 2a - 3b$
Thus,the relation is $2a - 3b = 1$.

(C) Since $\lim_{x \rightarrow 1} f(x)$ exists,the left-hand limit must equal the right-hand limit at $x = 1$.
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x)$
$1 + \frac{2(1)}{a} = a(1)$
$1 + \frac{2}{a} = a$
Multiplying by $a$ (assuming $a \neq 0$):
$a + 2 = a^2$
$a^2 - a - 2 = 0$
$(a - 2)(a + 1) = 0$
So,the possible values of $a$ are $a = 2$ and $a = -1$.
The sum of the cubes of these values is $(2)^3 + (-1)^3 = 8 - 1 = 7$.

(B) Given $\lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0$.
We can rewrite the limit as:
$\lim _{x \rightarrow 0} \left[ \frac{(a-n) n x-\tan x}{x} \right] \cdot \lim _{x \rightarrow 0} \frac{\sin n x}{x} = 0$.
Since $\lim _{x \rightarrow 0} \frac{\sin n x}{x} = n$,we have:
$n \cdot \lim _{x \rightarrow 0} \left[ (a-n) n - \frac{\tan x}{x} \right] = 0$.
$n [ (a-n) n - 1 ] = 0$.
Since $n > 0$,we must have $(a-n) n - 1 = 0$.
$(a-n) n = 1 \implies a-n = \frac{1}{n} \implies a = n + \frac{1}{n}$.
By $AM-GM$ inequality,for $n > 0$,$n + \frac{1}{n} \ge 2$.
The minimum value of $a$ is $2$ when $n = 1$.

(B) We are given the limit $L = \lim _{x \rightarrow 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$.
Using Taylor series expansions near $x=0$:
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
$e^x = 1 + x + \frac{x^2}{2!} + \dots$
Substituting these into the expression:
$(\cos x - 1) = -\frac{x^2}{2} + O(x^4)$
$(\cos x - e^x) = (1 - \frac{x^2}{2} + O(x^4)) - (1 + x + \frac{x^2}{2} + O(x^3)) = -x - x^2 + O(x^3)$
Thus,the numerator is $(-\frac{x^2}{2} + O(x^4))(-x - x^2 + O(x^3)) = \frac{x^3}{2} + O(x^4)$.
For the limit to be a finite non-zero real number,the power of $x$ in the denominator must match the lowest power of $x$ in the numerator.
Therefore,$n = 3$.

(D) Given,$\lim _{x \rightarrow 3} \frac{x^n - 3^n}{x - 3} = 108$.
Using the standard limit formula $\lim _{x \rightarrow a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1}$:
Here,$a = 3$,so $n \cdot 3^{n-1} = 108$.
$n \cdot 3^{n-1} = 108$.
Multiply both sides by $3$:
$n \cdot 3^n = 108 \times 3 = 324$.
We can write $324$ as $4 \times 81 = 4 \times 3^4$.
Comparing $n \cdot 3^n = 4 \times 3^4$,we get $n = 4$.
Thus,the value of $n$ is $4$.

(A) For the limit $\lim_{x \rightarrow 2} f(x)$ to exist,the Left Hand Limit $(LHL)$ must be equal to the Right Hand Limit $(RHL)$ at $x = 2$.
$LHL = \lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2} (4x - 5) = 4(2) - 5 = 3$.
$RHL = \lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2} (x - k) = 2 - k$.
Equating $LHL$ and $RHL$:
$3 = 2 - k$
$k = 2 - 3$
$k = -1$.
Thus,the correct option is $A$.

(D) Given that $\lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}=2$.
Simplifying the expression:
$\lim _{x \rightarrow \infty} \frac{x^3+1-(\alpha x+\beta)(x^2+1)}{x^2+1} = 2$
$\lim _{x \rightarrow \infty} \frac{x^3+1-(\alpha x^3+\alpha x+\beta x^2+\beta)}{x^2+1} = 2$
$\lim _{x \rightarrow \infty} \frac{(1-\alpha)x^3-\beta x^2-\alpha x+(1-\beta)}{x^2+1} = 2$.
For the limit to exist and be finite,the coefficient of $x^3$ must be $0$:
$1-\alpha = 0 \Rightarrow \alpha = 1$.
Now the limit becomes:
$\lim _{x \rightarrow \infty} \frac{-\beta x^2-x+(1-\beta)}{x^2+1} = 2$.
Dividing numerator and denominator by $x^2$:
$\lim _{x \rightarrow \infty} \frac{-\beta - \frac{1}{x} + \frac{1-\beta}{x^2}}{1 + \frac{1}{x^2}} = 2$.
As $x \rightarrow \infty$,the terms with $x$ in the denominator approach $0$:
$-\beta = 2 \Rightarrow \beta = -2$.
Thus,the ordered pair is $(\alpha, \beta) = (1, -2)$.

(D) Let $L = \lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8(P x^4-16)}{(x+\sqrt{x^2-2})^8+(x-\sqrt{x^2-2})^8} = 1$.
Divide the numerator and denominator by $x^4$ inside the terms or analyze the growth rates.
As $x \rightarrow \infty$,$\sqrt{2x+1} + \sqrt{2x-1} \approx 2\sqrt{2x}$ and $\sqrt{2x+1} - \sqrt{2x-1} = \frac{2}{\sqrt{2x+1} + \sqrt{2x-1}} \approx \frac{2}{2\sqrt{2x}} = \frac{1}{\sqrt{2x}}$.
Thus,$(\sqrt{2x+1} + \sqrt{2x-1})^8 \approx (2\sqrt{2x})^8 = 2^4 \cdot (2x)^4 = 16 \cdot 16x^4 = 256x^4$.
Also,$(x+\sqrt{x^2-2})^8 \approx (2x)^8 = 256x^8$.
The expression becomes $\lim _{x \rightarrow \infty} \frac{256x^4 + (\frac{1}{\sqrt{2x}})^8(Px^4-16)}{256x^8} = 1$.
Since $(\frac{1}{\sqrt{2x}})^8 = \frac{1}{16x^4}$,the expression is $\lim _{x \rightarrow \infty} \frac{256x^4 + \frac{1}{16x^4}(Px^4-16)}{256x^8} = 1$.
This simplifies to $\lim _{x \rightarrow \infty} \frac{256x^4 + \frac{P}{16} - \frac{1}{x^4}}{256x^8} = 1$.
For the limit to be $1$,the leading coefficients must match.
$256x^4 \cdot Px^4$ terms dominate,leading to $P = \frac{1}{16}$.

(C) Given $f(x) = \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}}$.
Let $t = \sqrt{\frac{x}{1-x}}$. Then $f(x) = t + \frac{1}{t}$.
We are given $\lim_{x \rightarrow m} (t + \frac{1}{t}) = 5/2$.
This implies $t + \frac{1}{t} = 5/2$,which simplifies to $2t^2 - 5t + 2 = 0$.
Solving for $t$: $(2t - 1)(t - 2) = 0$,so $t = 1/2$ or $t = 2$.
Case $1$: $\sqrt{\frac{m}{1-m}} = 1/2$ $\Rightarrow \frac{m}{1-m} = 1/4$ $\Rightarrow 4m = 1 - m$ $\Rightarrow 5m = 1$ $\Rightarrow m = 1/5$.
Case $2$: $\sqrt{\frac{m}{1-m}} = 2$ $\Rightarrow \frac{m}{1-m} = 4$ $\Rightarrow m = 4 - 4m$ $\Rightarrow 5m = 4$ $\Rightarrow m = 4/5$.
Thus,the set of possible values for $m$ is $\{1/5, 4/5\}$.

(C) We use the standard limit formula: $\lim _{x \rightarrow c} \frac{x^n - c^n}{x - c} = n c^{n-1}$.
Given $\lim _{x \rightarrow -a} \frac{x^7 - (-a)^7}{x - (-a)} = 7$.
Applying the formula with $n = 7$ and $c = -a$:
$7(-a)^{7-1} = 7$
$7(-a)^6 = 7$
$(-a)^6 = 1$
Since the exponent is even,$a^6 = 1$.
Taking the sixth root on both sides,$a = \pm 1$.

(B) For the limit to exist at $x=0$,the Left Hand Limit $(LHL)$ must equal the Right Hand Limit $(RHL)$.
$RHL = \lim_{x \rightarrow 0^+} \frac{2x+1}{x-2} = \frac{2(0)+1}{0-2} = -\frac{1}{2}$.
$LHL = \lim_{x \rightarrow 0^-} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}$.
Using rationalization:
$LHL = \lim_{x \rightarrow 0^-} \frac{(\sqrt{1+px} - \sqrt{1-px})(\sqrt{1+px} + \sqrt{1-px})}{x(\sqrt{1+px} + \sqrt{1-px})} = \lim_{x \rightarrow 0^-} \frac{(1+px) - (1-px)}{x(\sqrt{1+px} + \sqrt{1-px})} = \lim_{x \rightarrow 0^-} \frac{2px}{x(\sqrt{1+px} + \sqrt{1-px})} = \frac{2p}{1+1} = p$.
Equating $LHL = RHL$,we get $p = -\frac{1}{2}$.

(C) Given $\lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0, n \neq 0$.
Dividing the numerator and denominator by $x^2$,we get:
$\lim _{x \rightarrow 0} \left( \frac{(a-n) n x - \tan x}{x} \right) \left( \frac{\sin n x}{x} \right) = 0$.
$\lim _{x \rightarrow 0} \left( (a-n)n - \frac{\tan x}{x} \right) \left( n \cdot \frac{\sin n x}{n x} \right) = 0$.
Since $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin n x}{n x} = 1$,we have:
$((a-n)n - 1) \cdot n = 0$.
Since $n \neq 0$,we have $(a-n)n - 1 = 0$,which implies $an - n^2 = 1$,or $a = n + \frac{1}{n}$.
By the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$,for $n > 0$:
$\frac{n + \frac{1}{n}}{2} \geq \sqrt{n \cdot \frac{1}{n}} = 1$.
Therefore,$a \geq 2$.
The minimum possible positive value of $a$ is $2$.

(D) Given the quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$,we can write $ax^2+bx+c = a(x-\alpha)(x-\beta)$.
We need to evaluate the limit $L = \lim_{x \rightarrow \alpha} \frac{1-\cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2}$.
Using the identity $1-\cos(\theta) = 2\sin^2(\frac{\theta}{2})$,we get:
$L = \lim_{x \rightarrow \alpha} \frac{2\sin^2(\frac{a(x-\alpha)(x-\beta)}{2})}{(x-\alpha)^2}$.
Multiply and divide by $(\frac{a(x-\beta)}{2})^2$:
$L = \lim_{x}$ ${\rightarrow \alpha} 2 \left[ \frac{\sin(\frac{a(x-\alpha)(x-\beta)}{2})}{\frac{a(x-\alpha)(x-\beta)}{2}} \right]^2 \cdot \frac{a^2(x-\alpha)^2(x-\beta)^2}{4(x-\alpha)^2}$.
Since $\lim_{\theta \rightarrow 0} \frac{\sin(\theta)}{\theta} = 1$,the expression simplifies to:
$L = 2 \cdot 1^2 \cdot \frac{a^2(\alpha-\beta)^2}{4} = \frac{a^2(\alpha-\beta)^2}{2}$.

(A) Given the quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$,we can write $ax^2+bx+c = a(x-\alpha)(x-\beta)$.
Let $f(x) = ax^2+bx+c$. As $x \rightarrow \alpha$,$f(x) \rightarrow 0$.
Using the limit formula $\lim_{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2} = \frac{1}{2}$,we have:
$\lim_{x \rightarrow \alpha} \frac{1-\cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2} = \lim_{x \rightarrow \alpha} \frac{1-\cos(a(x-\alpha)(x-\beta))}{(a(x-\alpha)(x-\beta))^2} \times \frac{a^2(x-\alpha)^2(x-\beta)^2}{(x-\alpha)^2}$
$= \frac{1}{2} \times \lim_{x \rightarrow \alpha} a^2(x-\beta)^2$
$= \frac{1}{2} a^2(\alpha-\beta)^2$.

(D) Given $\lim _{x \rightarrow 0} \frac{e^x-a-\log (1+x)}{\sin x}=0$.
For the limit to exist and be finite,the numerator must approach $0$ as $x \rightarrow 0$ because the denominator $\sin x \rightarrow 0$.
Substituting $x=0$ in the numerator: $e^0 - a - \log(1+0) = 0$.
$1 - a - 0 = 0$.
Therefore,$a = 1$.
Checking with $a=1$: $\lim _{x \rightarrow 0} \frac{e^x-1-\log (1+x)}{\sin x}$.
Using $L$'Hopital's rule: $\lim _{x \rightarrow 0} \frac{e^x - \frac{1}{1+x}}{\cos x} = \frac{1-1}{1} = 0$.
Thus,the condition holds for $a=1$.

(A) Given $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+a^2\right)\right\}^{1 / x} = 2 a \sin ^2 \theta$.
Using the standard limit $\lim _{x \rightarrow 0} (1+f(x))^{1/x} = e^{\lim _{x \rightarrow 0} \frac{f(x)}{x}}$,we get:
$e^{\lim _{x \rightarrow 0} \frac{x \log(1+a^2)}{x}} = e^{\log(1+a^2)} = 1+a^2$.
Equating this to the given expression:
$1+a^2 = 2a \sin^2 \theta$.
Rearranging as a quadratic in $a$:
$a^2 - (2 \sin^2 \theta)a + 1 = 0$.
For $a$ to be a real number,the discriminant $D \ge 0$:
$D = (2 \sin^2 \theta)^2 - 4(1)(1) \ge 0$
$\Rightarrow 4 \sin^4 \theta - 4 \ge 0$
$\Rightarrow \sin^4 \theta \ge 1$.
Since $\sin^4 \theta \le 1$ for all $\theta$,the only possibility is $\sin^4 \theta = 1$.
$\Rightarrow \sin^2 \theta = 1 = \sin^2 \frac{\pi}{2}$.
$\Rightarrow \theta = n\pi \pm \frac{\pi}{2}, (n \in Z)$.

(D) Given $\lim _{x \rightarrow 2} \frac{3 x^2-a x+5 b}{x-2}=17$.
Since the limit exists and the denominator approaches $0$ as $x \rightarrow 2$,the numerator must also approach $0$ at $x=2$.
$3(2)^2 - a(2) + 5b = 0$ $\Rightarrow 12 - 2a + 5b = 0$ $\Rightarrow 2a - 5b = 12$.
Applying $L$'Hospital's rule:
$\lim _{x \rightarrow 2} \frac{d}{dx}(3x^2 - ax + 5b) / \frac{d}{dx}(x-2) = 17$.
$\lim _{x}$ ${\rightarrow 2} (6x - a) = 17$ $\Rightarrow 6(2) - a = 17$ $\Rightarrow 12 - a = 17$ $\Rightarrow a = -5$.
Substitute $a = -5$ into $2a - 5b = 12$:
$2(-5) - 5b = 12$ $\Rightarrow -10 - 5b = 12$ $\Rightarrow -5b = 22$ $\Rightarrow b = -\frac{22}{5}$.
Therefore,$ab = (-5) \times (-\frac{22}{5}) = 22$.

(D) Given $\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60} = \frac{11}{9}$.
Since the limit exists,the numerator must be $0$ at $x=4$: $2(16) + (3+2a)(4) + 3a = 0$.
$32 + 12 + 8a + 3a = 0 \Rightarrow 11a = -44 \Rightarrow a = -4$.
Alternatively,using $L$'Hospital's Rule: $\lim _{x \rightarrow 4} \frac{4x + 3 + 2a}{3x^2 - 4x - 23} = \frac{16 + 3 + 2a}{48 - 16 - 23} = \frac{19 + 2a}{9} = \frac{11}{9}$.
$19 + 2a = 11 \Rightarrow 2a = -8 \Rightarrow a = -4$.
Now,evaluate $\lim _{x \rightarrow -4} \frac{x^2+9x+20}{x^2-x-20} = \lim _{x \rightarrow -4} \frac{(x+4)(x+5)}{(x+4)(x-5)} = \lim _{x \rightarrow -4} \frac{x+5}{x-5}$.
Substituting $x = -4$: $\frac{-4+5}{-4-5} = \frac{1}{-9} = -\frac{1}{9}$.

(D) Let $f(x) = px^2 + qx + r$. Since $a$ and $b$ are roots,$f(x) = p(x - a)(x - b)$.
We need to evaluate $L = \lim_{x \rightarrow b} \frac{1 - \cos 2(f(x))}{2(px - pb)^2}$.
Using the identity $1 - \cos \theta = 2 \sin^2(\theta/2)$,we have $1 - \cos 2(f(x)) = 2 \sin^2(f(x))$.
So,$L = \lim_{x \rightarrow b} \frac{2 \sin^2(f(x))}{2 p^2(x - b)^2} = \frac{1}{p^2} \lim_{x \rightarrow b} \left( \frac{\sin(f(x))}{x - b} \right)^2$.
Since $f(x) = p(x - a)(x - b)$,we have $\frac{f(x)}{x - b} = p(x - a)$.
As $x \rightarrow b$,$p(x - a) \rightarrow p(b - a)$.
Thus,$L = \frac{1}{p^2} \lim_{x \rightarrow b} \left( \frac{\sin(p(x - a)(x - b))}{p(x - a)(x - b)} \cdot p(x - a) \right)^2$.
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we get $L = \frac{1}{p^2} \cdot (p(b - a))^2 = \frac{p^2(b - a)^2}{p^2} = (b - a)^2$.
Note: The provided options seem to have a typo. $(b - a)^2$ is equivalent to $a^2 - 2ab + b^2$,which is option $D$.

(C) Let $f(x) = \cos 4x + a \cos 2x + b$. For the limit to be finite as $x \rightarrow 0$,the numerator must approach $0$ as $x \rightarrow 0$.
Using the Taylor series expansion: $\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots$
$\cos 4x = 1 - \frac{(4x)^2}{2} + \frac{(4x)^4}{24} - \dots = 1 - 8x^2 + \frac{32}{3}x^4 - \dots$
$a \cos 2x = a(1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} - \dots) = a - 2ax^2 + \frac{2a}{3}x^4 - \dots$
Substituting these into the expression:
$f(x) = (1 + a + b) - (8 + 2a)x^2 + (\frac{32}{3} + \frac{2a}{3})x^4 + \dots$
For the limit to be finite,the coefficients of $x^0$ and $x^2$ must be zero.
$1 + a + b = 0$ and $8 + 2a = 0$.
From $8 + 2a = 0$,we get $a = -4$.
Substituting $a = -4$ into $1 + a + b = 0$,we get $1 - 4 + b = 0$,which gives $b = 3$.
Thus,the values are $a = -4$ and $b = 3$.

(D) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we can write $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.
We need to evaluate the limit $L = \lim_{x \rightarrow \beta} \frac{1 - \cos(a(x - \alpha)(x - \beta))}{(x - \beta)^2}$.
Using the identity $1 - \cos(\theta) = 2 \sin^2(\frac{\theta}{2})$,we get:
$L = \lim_{x \rightarrow \beta} \frac{2 \sin^2(\frac{a(x - \alpha)(x - \beta)}{2})}{(x - \beta)^2}$.
Multiplying and dividing by $(\frac{a(x - \alpha)}{2})^2$,we have:
$L = \lim_{x}$ ${\rightarrow \beta} 2 \left[ \frac{\sin(\frac{a(x - \alpha)(x - \beta)}{2})}{\frac{a(x - \alpha)(x - \beta)}{2}} \right]^2 \times \frac{a^2(x - \alpha)^2}{4}$.
As $x \rightarrow \beta$,the term in the bracket approaches $1$.
Thus,$L = 2 \times 1^2 \times \frac{a^2(\beta - \alpha)^2}{4} = \frac{a^2(\alpha - \beta)^2}{2}$.

(B) Given $\lim _{x \rightarrow \infty} \left(\frac{x^2+1}{x+1}-ax-b\right) = 0$.
Taking the common denominator:
$\lim _{x \rightarrow \infty} \frac{x^2+1-ax(x+1)-b(x+1)}{x+1} = 0$
$\lim _{x \rightarrow \infty} \frac{x^2+1-ax^2-ax-bx-b}{x+1} = 0$
$\lim _{x \rightarrow \infty} \frac{(1-a)x^2 - (a+b)x + (1-b)}{x+1} = 0$
For the limit to be $0$ as $x \rightarrow \infty$,the coefficient of the highest power of $x$ in the numerator must be $0$ and the degree of the numerator must be less than the degree of the denominator.
Thus,$1-a = 0 \implies a = 1$.
Substituting $a=1$ into the coefficient of $x$: $-(a+b) = 0 \implies -(1+b) = 0 \implies b = -1$.
Therefore,$a=1$ and $b=-1$.

(B) Given the limit: $\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x} = 1$.
Using the Taylor series expansion for $\sin x = x - \frac{x^3}{6} + \dots$ and $\sin 2x = 2x - \frac{8x^3}{6} + \dots$ and $\tan x = x + \frac{x^3}{3} + \dots$:
$\lim _{x \rightarrow 0} \frac{2 a(x - \frac{x^3}{6}) - (2x - \frac{8x^3}{6})}{x^3} = 1$.
$\lim _{x \rightarrow 0} \frac{(2a - 2)x + (\frac{8}{6} - \frac{2a}{6})x^3}{x^3} = 1$.
For the limit to exist,the coefficient of $x$ must be $0$,so $2a - 2 = 0$,which gives $a = 1$.
Substituting $a = 1$ into the expression: $\lim _{x \rightarrow 0} \frac{(8/6 - 2/6)x^3}{x^3} = \frac{6/6}{1} = 1$.
Thus,the value of $a$ is $1$.

(C) Using the Taylor series expansion for $x \rightarrow 0$:
$e^{(a-1)x} = 1 + (a-1)x + \frac{(a-1)^2 x^2}{2} + O(x^3)$
$2 \cos(bx) = 2(1 - \frac{b^2 x^2}{2} + O(x^4)) = 2 - b^2 x^2 + O(x^4)$
$(c-2)e^{-x} = (c-2)(1 - x + \frac{x^2}{2} + O(x^3))$
Denominator: $x(1 - \frac{x^2}{2}) - (x - \frac{x^2}{2} + \frac{x^3}{3}) = x - \frac{x^3}{2} - x + \frac{x^2}{2} - \frac{x^3}{3} = \frac{x^2}{2} - \frac{5x^3}{6} + O(x^4)$
For the limit to exist and equal $2$,the coefficients of $x^0$ and $x^1$ in the numerator must be $0$:
Constant term: $1 + 2 + c - 2 = c + 1 = 0 \Rightarrow c = -1$
Coefficient of $x$: $(a-1) - (c-2) = a - 1 - c + 2 = a - c + 1 = 0$ $\Rightarrow a - (-1) + 1 = 0$ $\Rightarrow a = -2$
Coefficient of $x^2$: $\frac{(a-1)^2}{2} - b^2 + \frac{c-2}{2} = 2 \times \frac{1}{2} = 1$
$\frac{(-2-1)^2}{2} - b^2 + \frac{-1-2}{2} = 1$ $\Rightarrow \frac{9}{2} - b^2 - \frac{3}{2} = 1$ $\Rightarrow 3 - b^2 = 1$ $\Rightarrow b^2 = 2$
Thus,$a^2 + b^2 + c^2 = (-2)^2 + 2 + (-1)^2 = 4 + 2 + 1 = 7$.

(C) Let $L = \lim_{x \to 0} \frac{1 - \cos(\alpha x) \cos((\alpha+1)x) \cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}$.
Using the approximation $\cos \theta \approx 1 - \frac{\theta^2}{2}$ and $\sin \theta \approx \theta$ as $x \to 0$:
$L = \lim_{x \to 0} \frac{1 - (1 - \frac{\alpha^2 x^2}{2})(1 - \frac{(\alpha+1)^2 x^2}{2})(1 - \frac{(\alpha+2)^2 x^2}{2})}{((\alpha+1)x)^2}$.
Neglecting higher order terms of $x^4$ and above,the numerator becomes:
$1 - (1 - \frac{x^2}{2}(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2)) = \frac{x^2}{2}(\alpha^2 + \alpha^2 + 2\alpha + 1 + \alpha^2 + 4\alpha + 4) = \frac{x^2}{2}(3\alpha^2 + 6\alpha + 5)$.
Thus,$L = \frac{3\alpha^2 + 6\alpha + 5}{2(\alpha+1)^2} = 2$.
$3\alpha^2 + 6\alpha + 5 = 4(\alpha^2 + 2\alpha + 1) = 4\alpha^2 + 8\alpha + 4$.
Rearranging gives $\alpha^2 + 2\alpha - 1 = 0$.
The product of the roots of this quadratic equation is given by $\frac{c}{a} = \frac{-1}{1} = -1$.

(B) For the limit to exist,the numerator must approach $0$ as $x \to 2$. Thus,$8 - 20 + 2a + b = 0 \Rightarrow 2a + b = 12$.
Let $t = x-2$,so $x = t+2$. As $x \to 2$,$t \to 0$.
The denominator becomes $(\sqrt{t+1}-1)\log_e(t+1) \approx (t/2)(t) = t^2/2$.
The numerator becomes $\sin((t+2)^3 - 5(t+2)^2 + a(t+2) + b) = \sin(t^3 + 6t^2 + 12t + 8 - 5t^2 - 20t - 20 + at + 2a + b) = \sin(t^3 + t^2 + (a-8)t + (2a+b-12))$.
Since $2a+b=12$,the expression simplifies to $\sin(t^3 + t^2 + (a-8)t)$.
For the limit to exist and be finite,the coefficient of $t$ must be $0$,so $a-8=0 \Rightarrow a=8$.
Substituting $a=8$ into $2a+b=12$,we get $16+b=12 \Rightarrow b=-4$.
The numerator is now $\sin(t^3+t^2) \approx t^2$ as $t \to 0$.
The limit is $\lim_{t \to 0} \frac{t^2}{t^2/2} = 2$,so $m=2$.
Finally,$a+b+m = 8 - 4 + 2 = 6$.