L'Hospital's rule and Limit of Indeterminate Form Questions in English

(B) Let the limit be $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_{2}^{\sec ^{2} x} f(t) dt}{x^{2}-\frac{\pi^{2}}{16}}$.
Since the form is $\frac{0}{0}$,we apply $L$'$H$ôpital's rule:
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \cdot f(\sec^2 x) \cdot \frac{d}{dx}(\sec^2 x)}{2x}$.
Using the chain rule,$\frac{d}{dx}(\sec^2 x) = 2 \sec x \cdot \sec x \tan x = 2 \sec^2 x \tan x$.
Substituting this back:
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \cdot f(\sec^2 x) \cdot 2 \sec^2 x \tan x}{2x}$.
At $x = \frac{\pi}{4}$,$\sec^2 x = 2$,$\tan x = 1$,and $x = \frac{\pi}{4}$.
$L = \frac{\frac{\pi}{4} \cdot f(2) \cdot 2 \cdot 2 \cdot 1}{2 \cdot \frac{\pi}{4}} = \frac{\pi \cdot f(2)}{\frac{\pi}{2}} = 2 f(2)$.

(B) Given $f(x) = x^{6} + 2x^{4} + x^{3} + 2x + 3$.
First,calculate $f(1) = 1^{6} + 2(1)^{4} + 1^{3} + 2(1) + 3 = 1 + 2 + 1 + 2 + 3 = 9$.
The limit is $\lim_{x \rightarrow 1} \frac{x^{n} f(1) - f(x)}{x - 1} = 44$.
Since the limit exists and the denominator approaches $0$,the numerator must also approach $0$ as $x \rightarrow 1$.
$1^{n} f(1) - f(1) = 9 - 9 = 0$.
Applying $L$'$H$ôpital's Rule:
$\lim_{x \rightarrow 1} \frac{n x^{n-1} f(1) - f'(x)}{1} = 44$.
Calculate $f'(x) = 6x^{5} + 8x^{3} + 3x^{2} + 2$.
At $x = 1$,$f'(1) = 6 + 8 + 3 + 2 = 19$.
Substituting these values: $n(1)^{n-1}(9) - 19 = 44$.
$9n - 19 = 44$.
$9n = 63$.
$n = 7$.

(D) Let $L = \lim _{x \rightarrow 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$.
Since $f(2)=4$,the expression becomes $\lim _{x \rightarrow 2} \frac{4x^{2}-4 f(x)}{x-2}$,which is of the form $\frac{0}{0}$ as $x \rightarrow 2$.
Applying $L$'Hopital's Rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 2} \frac{\frac{d}{dx}(x^{2} f(2)-4 f(x))}{\frac{d}{dx}(x-2)}$
$L = \lim _{x \rightarrow 2} \frac{2x f(2)-4 f^{\prime}(x)}{1}$
Substitute $f(2)=4$ and $f^{\prime}(2)=1$:
$L = 2(2)(4) - 4(1)$
$L = 16 - 4 = 12$.

(B) Given $g(x)=\int\limits_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t$.
Notice that the integrand is the derivative of the product $f(t) \sec t$,i.e.,$\frac{d}{dt}(f(t) \sec t) = f^{\prime}(t) \sec t + f(t) \sec t \tan t$.
Thus,$g(x) = \int\limits_{x}^{\pi / 4} d(f(t) \sec t) = [f(t) \sec t]_{x}^{\pi / 4}$.
$g(x) = f\left(\frac{\pi}{4}\right) \sec\left(\frac{\pi}{4}\right) - f(x) \sec x$.
Substituting the given values $f\left(\frac{\pi}{4}\right) = \sqrt{2}$ and $\sec\left(\frac{\pi}{4}\right) = \sqrt{2}$,we get $g(x) = \sqrt{2} \cdot \sqrt{2} - f(x) \sec x = 2 - \frac{f(x)}{\cos x}$.
Now,$\lim\limits_{x \rightarrow(\pi/2)^{-}} g(x) = \lim\limits_{x \rightarrow(\pi/2)^{-}} \left(2 - \frac{f(x)}{\cos x}\right)$.
Since $f(\pi/2) = 0$ and $\cos(\pi/2) = 0$,this is a $0/0$ form.
Applying $L$'Hopital's Rule: $\lim\limits_{x \rightarrow(\pi/2)^{-}} \frac{f(x)}{\cos x} = \lim\limits_{x \rightarrow(\pi/2)^{-}} \frac{f^{\prime}(x)}{-\sin x} = \frac{f^{\prime}(\pi/2)}{-\sin(\pi/2)} = \frac{1}{-1} = -1$.
Therefore,$\lim\limits_{x \rightarrow(\pi/2)^{-}} g(x) = 2 - (-1) = 3$.

(A) Let $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$. This is a $\frac{0}{0}$ form.
Applying $L'H\hat{o}pital$ rule:
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{-7(\cos x+\sin x)^{6}(-\sin x+\cos x)}{-2 \sqrt{2} \cos 2 x}$
Note that $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.
Substituting this into the expression:
$L = \lim _{x}$ ${\rightarrow \frac{\pi}{4}} \frac{-7(\cos x+\sin x)^{6}(\cos x - \sin x)}{-2 \sqrt{2} (\cos x - \sin x)(\cos x + \sin x)}$
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{7(\cos x+\sin x)^{5}}{2 \sqrt{2}}$
At $x = \frac{\pi}{4}$,$(\cos x + \sin x) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
$L = \frac{7(\sqrt{2})^{5}}{2 \sqrt{2}} = \frac{7 \times 4 \sqrt{2}}{2 \sqrt{2}} = \frac{28}{2} = 14$.

(A) We have,$\lim _{x \rightarrow \infty} x^2 \int _{0}^{x} e^{t^3-x^3} dt = \lim _{x \rightarrow \infty} x^2 e^{-x^3} \int _{0}^{x} e^{t^3} dt = \lim _{x \rightarrow \infty} \frac{x^2 \int _{0}^{x} e^{t^3} dt}{e^{x^3}}$.
Since this is an $\frac{\infty}{\infty}$ form,we apply $L'Hospital's$ rule:
$= \lim _{x \rightarrow \infty} \frac{\frac{d}{dx} (x^2 \int _{0}^{x} e^{t^3} dt)}{\frac{d}{dx} (e^{x^3})} = \lim _{x \rightarrow \infty} \frac{2x \int _{0}^{x} e^{t^3} dt + x^2 e^{x^3}}{3x^2 e^{x^3}} = \lim _{x \rightarrow \infty} \left( \frac{2 \int _{0}^{x} e^{t^3} dt}{3x e^{x^3}} + \frac{1}{3} \right)$.
Applying $L'Hospital's$ rule again to the first term:
$= \lim _{x \rightarrow \infty} \frac{2 e^{x^3}}{3(e^{x^3} + x \cdot 3x^2 e^{x^3})} + \frac{1}{3} = \lim _{x \rightarrow \infty} \frac{2 e^{x^3}}{3 e^{x^3}(1 + 3x^3)} + \frac{1}{3} = 0 + \frac{1}{3} = \frac{1}{3}$.

(D) Let $L = \lim _{x \rightarrow 2} \frac{a^x+a^{3-x}-\left(a^2+a\right)}{a^{3-x}-a^{x / 2}}$.
Since the limit is in the form $\frac{0}{0}$ at $x = 2$,we apply $L$'$H$ôpital's rule:
$L = \lim _{x \rightarrow 2} \frac{\frac{d}{dx}(a^x+a^{3-x}-a^2-a)}{\frac{d}{dx}(a^{3-x}-a^{x/2})}$
$L = \lim _{x \rightarrow 2} \frac{a^x \ln a - a^{3-x} \ln a}{-a^{3-x} \ln a - \frac{1}{2} a^{x/2} \ln a}$
Cancel $\ln a$ from the numerator and denominator:
$L = \lim _{x \rightarrow 2} \frac{a^x - a^{3-x}}{-a^{3-x} - \frac{1}{2} a^{x/2}}$
Substitute $x = 2$:
$L = \frac{a^2 - a^1}{-a^1 - \frac{1}{2} a^1} = \frac{a^2 - a}{-\frac{3}{2} a}$
$L = \frac{a(a-1)}{-\frac{3}{2} a} = -\frac{2}{3}(a-1) = \frac{2}{3}(1-a)$.

(A) Let $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{(x-\pi/2)^2}$.
Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule and use the Leibniz integral rule:
$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{d}{dx} \int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{2(x-\pi/2)}$
$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{0 - \cos((x^3)^{1/3}) \cdot \frac{d}{dx}(x^3)}{2(x-\pi/2)}$
$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos(x) \cdot 3x^2}{2(x-\pi/2)}$
Let $h = x - \frac{\pi}{2}$,then $x = h + \frac{\pi}{2}$. As $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$.
$L = \lim _{h \rightarrow 0} \frac{-\cos(h + \pi/2) \cdot 3(h + \pi/2)^2}{2h}$
$L = \lim _{h \rightarrow 0} \frac{\sin(h) \cdot 3(h + \pi/2)^2}{2h}$
Since $\lim _{h \rightarrow 0} \frac{\sin h}{h} = 1$,we get:
$L = 1 \cdot \frac{3(\pi/2)^2}{2} = \frac{3 \cdot \pi^2/4}{2} = \frac{3 \pi^2}{8}$.

(D) Let $t = |\sin x|$. As $x \rightarrow 0$,$t \rightarrow 0^+$.
The expression becomes $\lim _{t \rightarrow 0^+} \frac{e^{2t}-2t-1}{t^2} \times \frac{\sin^2 x}{x^2}$.
Since $\lim _{x \rightarrow 0} \frac{\sin^2 x}{x^2} = 1$,we evaluate $\lim _{t \rightarrow 0^+} \frac{e^{2t}-2t-1}{t^2}$.
Using the Taylor series expansion $e^{2t} = 1 + 2t + \frac{(2t)^2}{2!} + \frac{(2t)^3}{3!} + \dots = 1 + 2t + 2t^2 + \frac{4t^3}{3} + \dots$.
$\lim _{t \rightarrow 0^+} \frac{(1 + 2t + 2t^2 + \dots) - 2t - 1}{t^2} = \lim _{t \rightarrow 0^+} \frac{2t^2 + \dots}{t^2} = 2$.
Therefore,the limit is $2 \times 1 = 2$.

(A) Using $L$'Hopital's rule,we differentiate the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 1} \frac{\frac{d}{dx}((5 x+1)^{1 / 3}-(x+5)^{1 / 3})}{\frac{d}{dx}((2 x+3)^{1 / 2}-(x+4)^{1 / 2})}$
$= \lim _{x}$ ${\rightarrow 1} \frac{\frac{1}{3}(5 x+1)^{-2 / 3} \cdot 5 - \frac{1}{3}(x+5)^{-2 / 3}}{\frac{1}{2}(2 x+3)^{-1 / 2} \cdot 2 - \frac{1}{2}(x+4)^{-1 / 2}}$
$= \frac{\frac{5}{3}(6)^{-2 / 3} - \frac{1}{3}(6)^{-2 / 3}}{(5)^{-1 / 2} - \frac{1}{2}(5)^{-1 / 2}}$
$= \frac{\frac{4}{3} \cdot 6^{-2 / 3}}{\frac{1}{2} \cdot 5^{-1 / 2}} = \frac{8}{3} \cdot 6^{-2 / 3} \cdot \sqrt{5} = \frac{8 \sqrt{5}}{3 \cdot 6^{2 / 3}}$
Since $6^{2 / 3} = (2 \cdot 3)^{2 / 3} = 2^{2 / 3} \cdot 3^{2 / 3}$,the expression becomes $\frac{8 \sqrt{5}}{3^{1/3} \cdot 2^{2/3}}$.
Comparing with $\frac{m \sqrt{5}}{n(2 n)^{2 / 3}}$,we have $m=8$ and $n=3$.
Thus,$8m + 12n = 8(8) + 12(3) = 64 + 36 = 100$.

(B) Given $f(x) = \int_0^x (t + \sin(1 - e^t)) dt$. We need to find $\lim_{x \rightarrow 0} \frac{f(x)}{x^3}$.
Since $f(0) = 0$ and the denominator is $0$ at $x = 0$,we apply $L'H\text{ôpital's Rule}$:
$\lim_{x \rightarrow 0} \frac{f'(x)}{3x^2} = \lim_{x \rightarrow 0} \frac{x + \sin(1 - e^x)}{3x^2}$.
Applying $L'H\text{ôpital's Rule}$ again:
$\lim_{x \rightarrow 0} \frac{1 + \cos(1 - e^x) \cdot (-e^x)}{6x}$.
Using the Taylor expansion $e^x \approx 1 + x + \frac{x^2}{2}$ and $\cos(\theta) \approx 1 - \frac{\theta^2}{2}$:
$1 - e^x \approx -x - \frac{x^2}{2}$.
$\cos(1 - e^x) \approx 1 - \frac{(-x - \frac{x^2}{2})^2}{2} \approx 1 - \frac{x^2}{2}$.
Substituting back:
$\lim_{x \rightarrow 0} \frac{1 - (1 - \frac{x^2}{2})(1 + x + \frac{x^2}{2})}{6x} = \lim_{x \rightarrow 0} \frac{1 - (1 + x + \frac{x^2}{2} - \frac{x^2}{2})}{6x} = \lim_{x \rightarrow 0} \frac{-x}{6x} = -\frac{1}{6}$.

(A) Let $L = \operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$.
Using the identity $a^b = e^{b \ln a}$,we have $(1+2x)^{\frac{1}{2x}} = e^{\frac{\ln(1+2x)}{2x}}$.
$L = \operatorname{Lim}_{x \rightarrow 0} \frac{e - e^{\frac{\ln(1+2x)}{2x}}}{x} = e \operatorname{Lim}_{x \rightarrow 0} \frac{1 - e^{\frac{\ln(1+2x)}{2x} - 1}}{x}$.
Using the expansion $\ln(1+t) = t - \frac{t^2}{2} + \dots$,for $t=2x$,$\ln(1+2x) = 2x - \frac{(2x)^2}{2} + \dots = 2x - 2x^2 + \dots$.
Then $\frac{\ln(1+2x)}{2x} = 1 - x + \dots$.
So,$L = e \operatorname{Lim}_{x \rightarrow 0} \frac{1 - e^{-x}}{x} = e \operatorname{Lim}_{x \rightarrow 0} \frac{1 - (1 - x)}{x} = e \operatorname{Lim}_{x \rightarrow 0} \frac{x}{x} = e$.

(C) The function is $f(x) = |x-1|$. For $x < 1$,$f(x) = -(x-1) = -x+1$. The left-hand derivative at $x=1$ is $p = \frac{d}{dx}(-x+1) = -1$.
Given $\lim_{x \rightarrow 1^{+}} g(x) = p$,we have $\lim_{x \rightarrow 1^{+}} \frac{(x-1)^n}{\log \cos^m(x-1)} = -1$.
Let $h = x-1$. As $x \rightarrow 1^{+}$,$h \rightarrow 0^{+}$. The limit becomes $\lim_{h \rightarrow 0^{+}} \frac{h^n}{\log \cos^m h} = -1$.
Using the property $\log \cos^m h = m \log \cos h$,we have $\lim_{h \rightarrow 0^{+}} \frac{h^n}{m \log \cos h} = -1$.
Using $L$'$H$ôpital's rule,$\lim_{h \rightarrow 0^{+}} \frac{n h^{n-1}}{m (\frac{-\sin h}{\cos h})} = \lim_{h \rightarrow 0^{+}} \frac{n h^{n-1}}{-m \tan h} = -\frac{n}{m} \lim_{h \rightarrow 0^{+}} \frac{h^{n-1}}{\tan h} = -1$.
For this limit to be a non-zero constant,we must have $n-1 = 1$,so $n=2$. Then the limit is $-\frac{2}{m} \lim_{h \rightarrow 0^{+}} \frac{h}{\tan h} = -\frac{2}{m} (1) = -1$,which implies $m=2$.
Thus,$n=2$ and $m=2$.

(B) Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule.
Using the Leibniz integral rule,the derivative of the numerator is $\frac{d}{dx} \int_0^x \frac{t \ln (1+t)}{t^4+4} dt = \frac{x \ln (1+x)}{x^4+4}$.
The derivative of the denominator $x^3$ is $3x^2$.
Thus,the limit becomes $\lim _{x \rightarrow 0} \frac{x \ln (1+x)}{3x^2(x^4+4)}$.
Simplifying,we get $\lim _{x \rightarrow 0} \frac{1}{3} \cdot \frac{\ln (1+x)}{x} \cdot \frac{1}{x^4+4}$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x} = 1$,we have $\frac{1}{3} \cdot 1 \cdot \frac{1}{0^4+4} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$.

(C) We need to evaluate $\lim_{x \rightarrow \alpha^{+}} g(x)$.
Let $L = \lim_{x \rightarrow \alpha^{+}} \frac{2 \ln(\sqrt{x}-\sqrt{\alpha})}{\ln(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}$. This is a $\frac{0}{0}$ indeterminate form.
Using $L$'$H$ôpital's rule:
$L = 2 \lim_{x}$ ${\rightarrow \alpha^{+}} \frac{\frac{1}{\sqrt{x}-\sqrt{\alpha}} \cdot \frac{1}{2\sqrt{x}}}{\frac{1}{e^{\sqrt{x}}-e^{\sqrt{\alpha}}} \cdot e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}}$
$L = 2 \lim_{x \rightarrow \alpha^{+}} \frac{e^{\sqrt{x}}-e^{\sqrt{\alpha}}}{e^{\sqrt{x}}(\sqrt{x}-\sqrt{\alpha})}$
$L = \frac{2}{e^{\sqrt{\alpha}}} \lim_{x}$ ${\rightarrow \alpha^{+}} \frac{e^{\sqrt{x}}-e^{\sqrt{\alpha}}}{\sqrt{x}-\sqrt{\alpha}}$
Applying $L$'$H$ôpital's rule again to the limit part:
$L = \frac{2}{e^{\sqrt{\alpha}}} \lim_{x}$ ${\rightarrow \alpha^{+}} \frac{e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}} = \frac{2}{e^{\sqrt{\alpha}}} \cdot e^{\sqrt{\alpha}} = 2$.
Now,$\lim_{x}$ ${\rightarrow \alpha^{+}} f(g(x)) = f(2) = \sin\left(\frac{2\pi}{12}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} = 0.50$.

(B) The expression inside the summation is $\frac{2\tan \theta}{1 - \tan^2 \theta} = \tan(2\theta)$,where $\theta = \frac{x}{2^{r+1}}$.
Thus,the term is $\tan(2 \cdot \frac{x}{2^{r+1}}) = \tan(x/2^r)$.
However,the sum given is $\sum_{r=0}^n \tan(x/2^{r+1})$. Using the identity $\tan \theta = \cot \theta - 2\cot(2\theta)$,the sum telescopes to $f(x) = \lim_{n \rightarrow \infty} (\cot(x/2^{n+1}) - \cot x) = \frac{1}{x} - \cot x$.
Wait,re-evaluating the sum: $\sum_{r=0}^n \tan(x/2^{r+1}) = \frac{1}{x} - \cot x$ is incorrect. The standard identity is $\tan \theta = \cot \theta - 2\cot 2\theta$.
Actually,the sum $\sum_{r=0}^n \tan(x/2^{r+1})$ converges to $\frac{1}{x}$ as $n \to \infty$.
Thus $f(x) = \frac{1}{x} - \cot x$ is not correct. Let us use the limit $f(x) = \lim_{n \to \infty} \sum_{r=0}^n \tan(x/2^{r+1}) = x$.
Given the limit $\lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)}$,if $f(x) = x$,the expression is $0/0$. Applying $L$'Hopital's rule: $\lim_{x \to 0} \frac{e^x - e^{f(x)}f'(x)}{1 - f'(x)}$.
Since $f(x) = x$,$f'(x) = 1$,this leads to $e^0(1-1)/(1-1)$,which is indeterminate.
Re-evaluating the sum: $\sum_{r=0}^n \tan(x/2^{r+1}) = \frac{1}{x} - \cot x$. As $x \to 0$,$f(x) \approx x/3$.
Using the limit $\lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} = \lim_{u \to 0} \frac{e^u - 1}{u} = 1$ where $u = x - f(x)$.

(C) Let $L = \lim _{t \rightarrow 0}\left(\int_0^1(3 x+5)^t d x\right)^{\frac{1}{t}}$. This is of the form $1^{\infty}$.
Using the formula $\lim_{t \to 0} f(t)^{g(t)} = e^{\lim_{t \to 0} g(t)(f(t)-1)}$,we have:
$L = e^{\lim_{t \to 0} \frac{1}{t} \left( \int_0^1 (3x+5)^t dx - 1 \right)}$.
Evaluating the integral: $\int_0^1 (3x+5)^t dx = \left[ \frac{(3x+5)^{t+1}}{3(t+1)} \right]_0^1 = \frac{8^{t+1} - 5^{t+1}}{3(t+1)}$.
So,$L = e^{\lim_{t \to 0} \frac{1}{t} \left( \frac{8^{t+1} - 5^{t+1}}{3(t+1)} - 1 \right)} = e^{\lim_{t \to 0} \frac{8^{t+1} - 5^{t+1} - 3(t+1)}{3t(t+1)}}$.
Applying $L$'Hopital's rule as $t \to 0$: $\frac{8 \ln 8 - 5 \ln 5 - 3}{3}$.
This simplifies to $e^{\ln(8^{8/3}) - \ln(5^{5/3}) - 1} = e^{\ln(\frac{8^{8/3}}{5^{5/3}}) - 1} = \frac{1}{e} \cdot \frac{8^{8/3}}{5^{5/3}} = \frac{1}{e} \cdot \frac{8^{2/3} \cdot 8^2}{5^{2/3} \cdot 5} = \frac{64}{5e} \left( \frac{8}{5} \right)^{2/3}$.
Comparing with $\frac{\alpha}{5e} \left( \frac{8}{5} \right)^{2/3}$,we get $\alpha = 64$.

(B) Given the limit $\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x}{x^3} = 2$.
Using the $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \left(\frac{1}{1-x^2}\right) + \beta \cos x - \beta x \sin x}{3 x^2} = 2$.
Using the Taylor series expansion for $\frac{1}{1-x^2} = 1+x^2+x^4+\dots$,$\cos x = 1-\frac{x^2}{2}+\dots$,and $\sin x = x-\frac{x^3}{6}+\dots$:
$\lim _{x}$ ${\rightarrow 0} \frac{\frac{\alpha}{2}(1+x^2+x^4+\dots) + \beta(1-\frac{x^2}{2}+\dots) - \beta x(x-\frac{x^3}{6}+\dots)}{3 x^2} = 2$.
Grouping terms by powers of $x$:
$\lim _{x}$ ${\rightarrow 0} \frac{(\frac{\alpha}{2} + \beta) + x^2(\frac{\alpha}{2} - \frac{\beta}{2} - \beta) + O(x^4)}{3 x^2} = 2$.
For the limit to exist and equal $2$,the constant term must be zero: $\frac{\alpha}{2} + \beta = 0 \implies \alpha = -2\beta$.
Then the limit becomes $\lim _{x \rightarrow 0} \frac{x^2(\frac{\alpha}{2} - \frac{3\beta}{2})}{3 x^2} = \frac{\alpha - 3\beta}{6} = 2$.
Substituting $\alpha = -2\beta$: $\frac{-2\beta - 3\beta}{6} = 2 \implies -5\beta = 12 \implies \beta = -2.4$.
Then $\alpha = -2(-2.4) = 4.8$.
Thus,$\alpha + \beta = 4.8 - 2.4 = 2.4$.

(C) Given $e^{x_0}+x_0=0$.
$g(x) = \frac{3x(e^x+1) - \alpha(e^x+x)}{3(e^x+1)} = x - \frac{\alpha(e^x+x)}{3(e^x+1)}$.
Since $e^{x_0}+x_0=0$,we have $g(x_0) = x_0 - 0 = x_0$.
Thus,$g(x_0) + e^{x_0} = x_0 + e^{x_0} = 0$.
Using $L$'Hopital's rule,$\lim _{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| = |g'(x_0)|$.
$g'(x) = 1 - \frac{\alpha}{3} \left( \frac{(e^x+1)(e^x+1) - (e^x+x)e^x}{(e^x+1)^2} \right)$.
At $x=x_0$,$e^{x_0}+x_0=0$,so $g'(x_0) = 1 - \frac{\alpha}{3} \left( \frac{(e^{x_0}+1)^2 - 0}{(e^{x_0}+1)^2} \right) = 1 - \frac{\alpha}{3}$.
For $\alpha=3$,$|g'(x_0)| = |1 - \frac{3}{3}| = 0$.
Therefore,the statement for $\alpha=3$ is true.

(B) We are given the limit: $\lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a}$.
Since $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,substituting $x=a$ gives $\frac{g(a)f(a)-g(a)f(a)}{a-a} = \frac{0}{0}$,which is an indeterminate form.
Applying $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow a} \frac{\frac{d}{dx}[g(x) f(a)-g(a) f(x)]}{\frac{d}{dx}[x-a]}$
$= \lim _{x \rightarrow a} \frac{g^{\prime}(x) f(a)-g(a) f^{\prime}(x)}{1}$
$= g^{\prime}(a) f(a)-g(a) f^{\prime}(a)$
$= (2)(2)-(-1)(1)$
$= 4+1 = 5$.

(A) Given $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$.
First,find the derivative $f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$.
Evaluating at $x = 1$,$f'(1) = 30 - 56 + 30 - 63 + 6 = -53$.
Now,consider the limit $L = \lim_{\alpha \rightarrow 0} \frac{f(1-\alpha) - f(1)}{\alpha^3 + 3\alpha}$.
Rewrite the expression as $L = \lim_{\alpha \rightarrow 0} \left( \frac{f(1-\alpha) - f(1)}{-\alpha} \right) \times \left( \frac{-\alpha}{\alpha^3 + 3\alpha} \right)$.
Since $\lim_{\alpha \rightarrow 0} \frac{f(1-\alpha) - f(1)}{-\alpha} = f'(1) = -53$ and $\lim_{\alpha \rightarrow 0} \frac{-\alpha}{\alpha(\alpha^2 + 3)} = \lim_{\alpha \rightarrow 0} \frac{-1}{\alpha^2 + 3} = -\frac{1}{3}$.
Therefore,$L = (-53) \times (-1/3) = \frac{53}{3}$.

(A) Let $f(x) = \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$.
As $x \rightarrow 3$,the numerator becomes $(84-3)^{\frac{1}{4}}-3 = 81^{\frac{1}{4}}-3 = 3-3 = 0$ and the denominator becomes $3-3 = 0$.
This is a $\frac{0}{0}$ indeterminate form.
Using $L$'Hopital's Rule,we differentiate the numerator and denominator with respect to $x$:
$\frac{d}{dx} ((84-x)^{\frac{1}{4}}-3) = \frac{1}{4}(84-x)^{-\frac{3}{4}} \times (-1) = -\frac{1}{4}(84-x)^{-\frac{3}{4}}$.
$\frac{d}{dx} (x-3) = 1$.
Now,evaluate the limit as $x \rightarrow 3$:
$\lim _{x}$ ${\rightarrow 3} \frac{-\frac{1}{4}(84-x)^{-\frac{3}{4}}}{1} = -\frac{1}{4}(84-3)^{-\frac{3}{4}} = -\frac{1}{4}(81)^{-\frac{3}{4}}$.
Since $81 = 3^4$,we have $81^{-\frac{3}{4}} = (3^4)^{-\frac{3}{4}} = 3^{-3} = \frac{1}{27}$.
Therefore,the limit is $-\frac{1}{4} \times \frac{1}{27} = -\frac{1}{108}$.

(A) We are given the limit $L = \lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x}$.
Since the form is $\frac{0}{0}$,we can use the Taylor series expansion $e^u = 1 + u + \frac{u^2}{2} + \dots$.
$e^{\tan x} = 1 + \tan x + \frac{\tan^2 x}{2} + \dots$
$e^x = 1 + x + \frac{x^2}{2} + \dots$
Substituting these into the numerator:
$e^{\tan x} - e^x = (\tan x - x) + \frac{1}{2}(\tan^2 x - x^2) + \dots$
Now,the limit becomes $\lim _{x \rightarrow 0} \frac{(\tan x - x) + \frac{1}{2}(\tan^2 x - x^2)}{\tan x - x} = \lim _{x \rightarrow 0} [1 + \frac{1}{2} \frac{\tan^2 x - x^2}{\tan x - x}]$.
Using $\tan x \approx x + \frac{x^3}{3}$,we have $\tan^2 x - x^2 \approx (x + \frac{x^3}{3})^2 - x^2 \approx x^2 + \frac{2x^4}{3} - x^2 = \frac{2x^4}{3}$.
Also,$\tan x - x \approx \frac{x^3}{3}$.
Thus,$\frac{\tan^2 x - x^2}{\tan x - x} \approx \frac{2x^4/3}{x^3/3} = 2x$.
As $x \rightarrow 0$,this term approaches $0$.
Therefore,the limit is $1 + 0 = 1$.

(D) Let $f(x) = x + 3x^2 + 5x^3 + 7x^4 - 166$.
First,check the value of $f(2)$:
$f(2) = 2 + 3(2^2) + 5(2^3) + 7(2^4) - 166 = 2 + 12 + 40 + 112 - 166 = 166 - 166 = 0$.
Since the limit is of the form $\frac{0}{0}$,we apply $L$'Hopital's Rule:
$\lim _{x \rightarrow 2} \frac{f'(x)}{1} = f'(2)$.
$f'(x) = \frac{d}{dx}(x + 3x^2 + 5x^3 + 7x^4 - 166) = 1 + 6x + 15x^2 + 28x^3$.
Now,evaluate $f'(2)$:
$f'(2) = 1 + 6(2) + 15(2^2) + 28(2^3) = 1 + 12 + 15(4) + 28(8) = 1 + 12 + 60 + 224 = 297$.
Thus,the limit is $297$.

(C) Let $L = \lim _{x \rightarrow 0^+} ((\sin x)^{\frac{1}{x}} + (\frac{1}{x})^{\sin x})$.
First,consider $L_1 = \lim _{x \rightarrow 0^+} (\sin x)^{\frac{1}{x}}$. As $x \rightarrow 0^+$,$\sin x \rightarrow 0^+$ and $\frac{1}{x} \rightarrow \infty$. Since $0$ raised to any positive power is $0$,$L_1 = 0$.
Now,consider $L_2 = \lim _{x \rightarrow 0^+} (\frac{1}{x})^{\sin x}$.
Taking the natural logarithm on both sides: $\ln L_2 = \lim _{x \rightarrow 0^+} \sin x \ln(\frac{1}{x}) = \lim _{x \rightarrow 0^+} -\sin x \ln x$.
This is an indeterminate form of type $0 \times \infty$. We rewrite it as: $\ln L_2 = -\lim _{x \rightarrow 0^+} \frac{\ln x}{\csc x}$.
Applying $L'H\hat{o}pital's$ Rule: $\ln L_2 = -\lim _{x \rightarrow 0^+} \frac{1/x}{-\csc x \cot x} = \lim _{x \rightarrow 0^+} \frac{\sin x \tan x}{x} = \lim _{x \rightarrow 0^+} (\frac{\sin x}{x}) \cdot \tan x$.
Since $\lim _{x \rightarrow 0^+} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0^+} \tan x = 0$,we have $\ln L_2 = 1 \times 0 = 0$.
Thus,$L_2 = e^0 = 1$.
Therefore,$L = L_1 + L_2 = 0 + 1 = 1$.

(C) Let $L = \lim _{x \rightarrow 2}(x-1)^{\frac{1}{3x-6}}$.
Since the form is $1^{\infty}$,we use the formula $\lim _{x \rightarrow a} f(x)^{g(x)} = e^{\lim _{x \rightarrow a} g(x)(f(x)-1)}$.
Here,$f(x) = x-1$ and $g(x) = \frac{1}{3x-6}$.
$L = e^{\lim _{x \rightarrow 2} \frac{1}{3(x-2)} (x-1-1)}$.
$L = e^{\lim _{x \rightarrow 2} \frac{x-2}{3(x-2)}}$.
$L = e^{\lim _{x \rightarrow 2} \frac{1}{3}} = e^{\frac{1}{3}}$.

(B) Let $L = \lim _{x \rightarrow 1} \frac{a b^x - a^x b}{x^2 - 1}$.
Since the limit is of the form $\frac{0}{0}$ at $x = 1$,we apply $L$' Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(a b^x - a^x b)}{\frac{d}{dx}(x^2 - 1)}$
$L = \lim _{x \rightarrow 1} \frac{a b^x \ln b - b a^x \ln a}{2x}$
Substituting $x = 1$:
$L = \frac{a b^1 \ln b - b a^1 \ln a}{2(1)}$
$L = \frac{ab \ln b - ab \ln a}{2}$
$L = \frac{ab}{2} (\ln b - \ln a)$
$L = \frac{ab}{2} \ln \left(\frac{b}{a}\right)$

(A) Given the limit: $\lim _{x \rightarrow 1} \left[\frac{\sqrt{x}-1}{\log x}\right]$
Since the form is $\frac{0}{0}$ as $x \rightarrow 1$,we apply $L'\text{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:
$= \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(\sqrt{x}-1)}{\frac{d}{dx}(\log x)}$
$= \lim _{x \rightarrow 1} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{x}}$
$= \lim _{x \rightarrow 1} \frac{x}{2\sqrt{x}} = \lim _{x \rightarrow 1} \frac{\sqrt{x}}{2}$
$= \frac{\sqrt{1}}{2} = \frac{1}{2}$

(D) $\lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x)$
$\Rightarrow \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \left(\frac{1-\sin x}{\cos x}\right)$ (which is of the $\frac{0}{0}$ form)
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$\Rightarrow \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \frac{\frac{d}{dx}(1-\sin x)}{\frac{d}{dx}(\cos x)}$
$\Rightarrow \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \frac{-\cos x}{-\sin x}$
$\Rightarrow \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \cot x$
Substituting $x = \frac{\pi}{2}$:
$\Rightarrow \cot \frac{\pi}{2} = 0$

(C) The given limit is $\lim _{x \rightarrow 0} \left( \frac{3^{x}-1}{x} \right)$,which is in the $\left( \frac{0}{0} \right)$ indeterminate form.
Applying $L'\text{Hospital's rule}$ by differentiating the numerator and the denominator with respect to $x$:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(3^{x}-1)}{\frac{d}{dx}(x)}$
$= \lim _{x \rightarrow 0} \frac{3^{x} \log 3}{1}$
$= 3^{0} \log 3 = 1 \cdot \log 3 = \log 3$.

(A) Let $L = \lim _{x \rightarrow 1} (\log _3 3x)^{\log _x 8}$.
This is an indeterminate form of type $1^{\infty}$.
We use the formula $\lim _{x \rightarrow a} f(x)^{g(x)} = e^{\lim _{x \rightarrow a} g(x)(f(x)-1)}$.
Here,$f(x) = \log _3 3x = \log _3 3 + \log _3 x = 1 + \log _3 x$ and $g(x) = \log _x 8$.
So,$L = \exp(\lim _{x \rightarrow 1} \log _x 8 (1 + \log _3 x - 1)) = \exp(\lim _{x \rightarrow 1} \log _x 8 \cdot \log _3 x)$.
Using the change of base formula,$\log _x 8 = \frac{\ln 8}{\ln x}$ and $\log _3 x = \frac{\ln x}{\ln 3}$.
Thus,$L = \exp(\lim _{x \rightarrow 1} \frac{\ln 8}{\ln x} \cdot \frac{\ln x}{\ln 3}) = \exp(\frac{\ln 8}{\ln 3}) = \exp(\log _3 8) = 8$.

(B) Let $L = \lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$.
Since the limit is in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital's$ rule:
$L = \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(x^4-\sqrt{x})}{\frac{d}{dx}(\sqrt{x}-1)}$
$L = \lim _{x \rightarrow 1} \frac{4x^3 - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$
$L = \lim _{x}$ ${\rightarrow 1} \left( \frac{4x^3}{\frac{1}{2\sqrt{x}}} - \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}} \right)$
$L = \lim _{x \rightarrow 1} (8x^3\sqrt{x} - 1)$
$L = 8(1)^3\sqrt{1} - 1 = 8 - 1 = 7$.

(C) Given limit: $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$
Substituting $x = \frac{\pi}{4}$,we get the form $\frac{\sqrt{2}(\frac{1}{\sqrt{2}})-1}{1-1} = \frac{0}{0}$.
Applying $L'\text{Hospital rule}$ by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{d}{dx}(\sqrt{2} \cos x - 1)}{\frac{d}{dx}(\cot x - 1)}$
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\operatorname{cosec}^2 x} = \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{\operatorname{cosec}^2 x}$
$L = \lim _{x \rightarrow \frac{\pi}{4}} \sqrt{2} \sin x \sin^2 x = \lim _{x \rightarrow \frac{\pi}{4}} \sqrt{2} \sin^3 x$
$L = \sqrt{2} \times (\frac{1}{\sqrt{2}})^3 = \sqrt{2} \times \frac{1}{2\sqrt{2}} = \frac{1}{2}$.

(C) Given limit: $\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}$
This is in the indeterminate form $\frac{0}{0}$.
Applying $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\log _{e}(1+x))}{\frac{d}{dx}(3^{x}-1)}$
$= \lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{3^{x} \cdot \log _{e} 3}$
Substituting $x = 0$:
$= \frac{\frac{1}{1+0}}{3^{0} \cdot \log _{e} 3} = \frac{1}{1 \cdot \log _{e} 3} = \frac{1}{\log _{e} 3}$
Using the property $\frac{1}{\log _{a} b} = \log _{b} a$,we get:
$= \log _{3} e$

(A) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(2^{x}-1)}{1-\cos x}$
Since this is a $\frac{0}{0}$ form,we apply $L$'Hopital's rule:
$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(x 2^{x}-x)}{\frac{d}{dx}(1-\cos x)}$
$L = \lim _{x \rightarrow 0} \frac{2^{x} + x 2^{x} \ln 2 - 1}{\sin x}$
This is still a $\frac{0}{0}$ form. Applying $L$'Hopital's rule again:
$L = \lim _{x \rightarrow 0} \frac{2^{x} \ln 2 + 2^{x} \ln 2 + x 2^{x} (\ln 2)^{2}}{\cos x}$
Substituting $x = 0$:
$L = \frac{2^{0} \ln 2 + 2^{0} \ln 2 + 0}{\cos 0} = \frac{\ln 2 + \ln 2}{1} = 2 \ln 2$
Thus,the correct option is $A$.

(A) Given limit is $\lim _{x \rightarrow 1} \frac{\tan \left(x^{2}-1\right)}{x-1}$.
Substituting $x=1$,we get the $\frac{0}{0}$ indeterminate form.
Using $L$'Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 1} \frac{\frac{d}{dx} \tan \left(x^{2}-1\right)}{\frac{d}{dx} (x-1)}$
$= \lim _{x \rightarrow 1} \frac{\sec ^{2}\left(x^{2}-1\right) \cdot (2x)}{1}$
$= \sec ^{2}(1^{2}-1) \cdot 2(1)$
$= \sec ^{2}(0) \cdot 2$
$= 1 \cdot 2 = 2$.

(C) Given the limit $ \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} $.
Substituting $ x = 0 $,we get the indeterminate form $ \left(\frac{0}{0}\right) $.
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $ x $:
$ \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(x^{2})} = \lim _{x \rightarrow 0} \frac{\sin x}{2x} $.
Using the standard limit $ \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 $,we get:
$ \frac{1}{2} \times \lim _{x \rightarrow 0} \frac{\sin x}{x} = \frac{1}{2} \times 1 = \frac{1}{2} $.
Thus,the value of the limit is $ \frac{1}{2} $.

(D) Let $L = \lim _{x \rightarrow -9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}$.
Since the form is $\frac{0}{0}$,we apply $L$'Hopital's Rule:
$L = \lim _{x \rightarrow -9} \frac{\frac{d}{dx}((2.5)^{81-x^2}) - \frac{d}{dx}((0.4)^{x+9})}{\frac{d}{dx}(x+9)}$
$L = \lim _{x \rightarrow -9} \frac{(2.5)^{81-x^2} \cdot \ln(2.5) \cdot (-2x) - (0.4)^{x+9} \cdot \ln(0.4) \cdot 1}{1}$
Substitute $x = -9$:
$L = (2.5)^{81-81} \cdot \ln(2.5) \cdot (-2(-9)) - (0.4)^{0} \cdot \ln(0.4)$
$L = (2.5)^0 \cdot \ln(2.5) \cdot 18 - 1 \cdot \ln(0.4)$
$L = 18 \ln(2.5) - \ln(0.4)$.
Since $\ln(0.4) = \ln(2.5^{-1}) = -\ln(2.5)$,we have $L = 18 \ln(2.5) + \ln(2.5) = 19 \ln(2.5)$.
However,checking the options provided,the expression $-19 \log(0.4)$ is equivalent to $-19 \ln(0.4) / \ln(10) = -19 \ln(2.5^{-1}) / \ln(10) = 19 \ln(2.5) / \ln(10) = 19 \log(2.5)$.
Given the structure,option $D$ is the intended answer.

(C) To evaluate the limit $\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{2}\right)^{5 / 7}-1}{x}$,we observe that it is in the indeterminate form $\frac{0}{0}$.
We can apply $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $x$:
$\lim _{x \rightarrow 0} \frac{\frac{d}{dx}\left(\left(1+\frac{x}{2}\right)^{5 / 7}-1\right)}{\frac{d}{dx}(x)}$
$= \lim _{x}$ ${\rightarrow 0} \frac{\frac{5}{7}\left(1+\frac{x}{2}\right)^{5 / 7 - 1} \cdot \frac{d}{dx}\left(1+\frac{x}{2}\right)}{1}$
$= \lim _{x \rightarrow 0} \frac{5}{7}\left(1+\frac{x}{2}\right)^{-2 / 7} \cdot \frac{1}{2}$
$= \frac{5}{7} \cdot (1+0)^{-2 / 7} \cdot \frac{1}{2}$
$= \frac{5}{14}$
Thus,the correct option is $C$.

(A) Let $f(x) = x+2 \sin x+3 \tan x-\tan ^3 x$ and $g(x) = \sqrt{x^2+2 \sin x+\tan x+3}-\sqrt{\sin ^2 x-2 \tan x-x+3}$.
As $x \rightarrow 0$,$f(0) = 0+0+0-0 = 0$ and $g(0) = \sqrt{3}-\sqrt{3} = 0$. This is a $\frac{0}{0}$ form.
We apply $L$'Hopital's rule by differentiating the numerator and denominator.
Numerator derivative: $f'(x) = 1+2 \cos x+3 \sec ^2 x-3 \tan ^2 x \sec ^2 x$.
At $x=0$,$f'(0) = 1+2(1)+3(1)-0 = 6$.
Denominator derivative: $g'(x) = \frac{2x+2 \cos x+\sec ^2 x}{2 \sqrt{x^2+2 \sin x+\tan x+3}} - \frac{2 \sin x \cos x-2 \sec ^2 x-1}{2 \sqrt{\sin ^2 x-2 \tan x-x+3}}$.
At $x=0$,$g'(0) = \frac{0+2+1}{2 \sqrt{3}} - \frac{0-2-1}{2 \sqrt{3}} = \frac{3}{2 \sqrt{3}} - \frac{-3}{2 \sqrt{3}} = \frac{6}{2 \sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
The limit is $\frac{f'(0)}{g'(0)} = \frac{6}{\sqrt{3}} = 2 \sqrt{3}$.

(A) Let $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}$. This is a $\frac{0}{0}$ form.
Applying $L'H\hat{o}pital's$ Rule:
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{-5(\cos x+\sin x)^4(-\sin x+\cos x)}{-2 \cos 2 x}$.
Using the identity $1-\sin 2x = (\cos x - \sin x)^2$ and $\cos 2x = (\cos x - \sin x)(\cos x + \sin x)$:
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{5(\cos x+\sin x)^4(\cos x-\sin x)}{2(\cos x-\sin x)(\cos x+\sin x)}$.
$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{5}{2}(\cos x+\sin x)^3$.
Substituting $x = \frac{\pi}{4}$:
$L = \frac{5}{2}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})^3 = \frac{5}{2}(\sqrt{2})^3 = \frac{5}{2}(2 \sqrt{2}) = 5 \sqrt{2}$.

(A) Given limit is $\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8}$.
Since it is a $\frac{0}{0}$ indeterminate form,we apply $L'H\hat{o}pital's$ rule by differentiating the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 2} \frac{\frac{d}{dx}(\sqrt{1+4 x}-\sqrt{3+3 x})}{\frac{d}{dx}(x^3-8)} = \lim _{x \rightarrow 2} \frac{\frac{4}{2\sqrt{1+4x}} - \frac{3}{2\sqrt{3+3x}}}{3x^2}$.
Substituting $x = 2$:
$= \frac{\frac{4}{2\sqrt{9}} - \frac{3}{2\sqrt{9}}}{3(2^2)} = \frac{\frac{4}{6} - \frac{3}{6}}{12} = \frac{\frac{1}{6}}{12} = \frac{1}{72}$.

(A) Using $L'H\hat{o}pital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\cos 2x - \cos 3x)}{\frac{d}{dx}(\cos 4x - \cos 5x)} = \lim _{x \rightarrow 0} \frac{-2 \sin 2x + 3 \sin 3x}{-4 \sin 4x + 5 \sin 5x}$
Since this is still in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital's$ rule again:
$\lim _{x \rightarrow 0} \frac{-4 \cos 2x + 9 \cos 3x}{-16 \cos 4x + 25 \cos 5x}$
Substituting $x = 0$:
$= \frac{-4 \cos(0) + 9 \cos(0)}{-16 \cos(0) + 25 \cos(0)} = \frac{-4(1) + 9(1)}{-16(1) + 25(1)} = \frac{5}{9}$

(B) Let $L = \lim _{x \rightarrow \pi / 6} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}$.
Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $x$:
$L = \lim _{x \rightarrow \pi / 6} \frac{\frac{d}{dx}(3 \sin x-\sqrt{3} \cos x)}{\frac{d}{dx}(6 x-\pi)}$
$L = \lim _{x \rightarrow \pi / 6} \frac{3 \cos x+\sqrt{3} \sin x}{6}$
Substituting $x = \frac{\pi}{6}$:
$L = \frac{3 \cos(\pi/6) + \sqrt{3} \sin(\pi/6)}{6}$
$L = \frac{3(\frac{\sqrt{3}}{2}) + \sqrt{3}(\frac{1}{2})}{6}$
$L = \frac{\frac{3\sqrt{3} + \sqrt{3}}{2}}{6} = \frac{4\sqrt{3}}{12} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

(D) Let $L = \lim _{x \rightarrow 1}(1-x) \tan \left(\frac{\pi x}{2}\right)$.
This is an indeterminate form of type $0 \times \infty$.
We can rewrite the limit as:
$L = \lim _{x \rightarrow 1} \frac{1-x}{\cot \left(\frac{\pi x}{2}\right)}$.
Applying $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(1-x)}{\frac{d}{dx}(\cot \left(\frac{\pi x}{2}\right))}$
$L = \lim _{x \rightarrow 1} \frac{-1}{-\csc^2 \left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}}$
$L = \frac{1}{\frac{\pi}{2} \csc^2 \left(\frac{\pi}{2}\right)}$
Since $\csc \left(\frac{\pi}{2}\right) = 1$,we get:
$L = \frac{1}{\frac{\pi}{2} \cdot 1^2} = \frac{2}{\pi}$.
Thus,the correct option is $D$.

(B) Given $\alpha = \lim_{x \rightarrow 0} \frac{x(2^x - 1)}{1 - \cos x}$.
Applying $L'H\hat{o}pital's$ rule:
$\alpha = \lim_{x \rightarrow 0} \frac{2^x - 1 + x \cdot 2^x \ln 2}{\sin x}$.
Applying $L'H\hat{o}pital's$ rule again:
$\alpha = \lim_{x \rightarrow 0} \frac{2^x \ln 2 + 2^x \ln 2 + x \cdot 2^x (\ln 2)^2}{\cos x} = \frac{\ln 2 + \ln 2 + 0}{1} = 2 \ln 2$.
Now,$\beta = \lim_{x \rightarrow 0} \frac{x(2^x - 1)}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}$.
Rationalizing the denominator:
$\beta = \lim_{x \rightarrow 0} \frac{x(2^x - 1)(\sqrt{1 + x^2} + \sqrt{1 - x^2})}{(1 + x^2) - (1 - x^2)} = \lim_{x \rightarrow 0} \frac{x(2^x - 1)(\sqrt{1 + x^2} + \sqrt{1 - x^2})}{2x^2}$.
$\beta = \frac{1}{2} \lim_{x \rightarrow 0} \frac{2^x - 1}{x} \cdot (\sqrt{1 + x^2} + \sqrt{1 - x^2}) = \frac{1}{2} \cdot \ln 2 \cdot (1 + 1) = \ln 2$.
Thus,$\alpha = 2 \beta$.

(D) Given the limit $L = \lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}$.
Since $f(9)=9$,the expression takes the form $\frac{\sqrt{9}-3}{\sqrt{9}-3} = \frac{0}{0}$,which is an indeterminate form.
Applying $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 9} \frac{\frac{d}{dx}(\sqrt{f(x)}-3)}{\frac{d}{dx}(\sqrt{x}-3)}$
$L = \lim _{x \rightarrow 9} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}}$
$L = \lim _{x \rightarrow 9} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$
Substituting $x=9$:
$L = \frac{f'(9) \cdot \sqrt{9}}{\sqrt{f(9)}} = \frac{4 \cdot 3}{\sqrt{9}} = \frac{12}{3} = 4$.

(C) Let $\ell = \lim _{x \rightarrow \frac{\pi}{2}} \frac{2x-\pi}{\cos x}$.
This is an indeterminate form of type $\frac{0}{0}$.
Applying $L$' Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$\ell = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{d}{dx}(2x-\pi)}{\frac{d}{dx}(\cos x)}$
$\ell = \lim _{x \rightarrow \frac{\pi}{2}} \frac{2}{-\sin x}$
Substituting $x = \frac{\pi}{2}$:
$\ell = \frac{2}{-\sin(\frac{\pi}{2})} = \frac{2}{-1} = -2$.

(D) Given limit: $L = \lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = \lim _{x \rightarrow 0} \left[ \frac{e^x - 1 - x}{x(e^x - 1)} \right]$
This is in the $\frac{0}{0}$ indeterminate form.
Applying $L'\text{Hopital's rule}$:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(e^x - 1 - x)}{\frac{d}{dx}(xe^x - x)} = \lim _{x \rightarrow 0} \frac{e^x - 1}{e^x + xe^x - 1}$
This is still in the $\frac{0}{0}$ form.
Applying $L'\text{Hopital's rule}$ again:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(e^x + xe^x - 1)} = \lim _{x \rightarrow 0} \frac{e^x}{e^x + e^x + xe^x} = \lim _{x \rightarrow 0} \frac{e^x}{2e^x + xe^x}$
Substituting $x = 0$:
$= \frac{e^0}{2e^0 + 0 \cdot e^0} = \frac{1}{2 + 0} = \frac{1}{2}$

(B) Given the limit: $L = \lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$
We can rewrite the expression as: $L = \lim _{x \rightarrow 0} \frac{e^{\sin x}(e^{x-\sin x}-1)}{2(x-\sin x)}$
Using the standard limit $\lim _{u \rightarrow 0} \frac{e^u-1}{u} = 1$,where $u = x - \sin x$:
As $x \rightarrow 0$,$u = x - \sin x \rightarrow 0$.
Therefore,the limit becomes: $L = \lim _{x \rightarrow 0} \frac{e^{\sin x}}{2} \times \lim _{u \rightarrow 0} \frac{e^u-1}{u}$
$L = \frac{e^0}{2} \times 1 = \frac{1}{2} \times 1 = \frac{1}{2}$