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

(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: $L = \lim _{x \rightarrow 0} \frac{e^{\sin x}}{2} \times \lim _{u \rightarrow 0} \frac{e^u-1}{u}$
$L = \frac{e^0}{2} \times 1 = \frac{1}{2} \times 1 = \frac{1}{2}$

(B) Let $L = \lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}$.
Since $\lim _{x \rightarrow 0} \frac{e^x-1}{x} = 1$ and the exponent $\frac{x}{x+1-e^x} \to \infty$ as $x \to 0$,this is an indeterminate form of type $1^\infty$.
We use the formula $\lim _{x \to a} f(x)^{g(x)} = e^{\lim _{x \to a} g(x)(f(x)-1)}$.
$L = e^{\lim _{x \to 0} \left( \frac{x}{x+1-e^x} \right) \left( \frac{e^x-1}{x} - 1 \right)}$.
$L = e^{\lim _{x \to 0} \left( \frac{x}{x+1-e^x} \right) \left( \frac{e^x-1-x}{x} \right)}$.
$L = e^{\lim _{x \to 0} \frac{e^x-1-x}{x+1-e^x}}$.
$L = e^{\lim _{x \to 0} \frac{-(x+1-e^x)}{x+1-e^x}} = e^{-1}$.

(C) Given the limit $L = \lim_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$.
Substituting $x = 0$ gives the $\frac{0}{0}$ indeterminate form.
Applying $L$'Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
$L = \lim_{x \to 0} \frac{2f'(x) - 3 \cdot 2f'(2x) + 4f'(4x)}{2x} = \lim_{x \to 0} \frac{2f'(x) - 6f'(2x) + 4f'(4x)}{2x}$.
Applying $L$'Hospital's rule again:
$L = \lim_{x \to 0} \frac{2f''(x) - 6 \cdot 2f''(2x) + 4 \cdot 4f''(4x)}{2} = \lim_{x \to 0} \frac{2f''(x) - 12f''(2x) + 16f''(4x)}{2}$.
Since $f''(x)$ is continuous at $x = 0$,we have $\lim_{x \to 0} f''(x) = f''(0) = 4$.
Substituting $x = 0$ into the expression:
$L = \frac{2f''(0) - 12f''(0) + 16f''(0)}{2} = \frac{6f''(0)}{2} = 3f''(0)$.
Given $f''(0) = 4$,we get $L = 3 \times 4 = 12$.

(D) Given the polynomial equation $x^n+px+q=0$ with roots $\alpha_1, \alpha_2, \ldots, \alpha_n$,we can write it as:
$x^n+px+q = (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n)$.
Dividing both sides by $(x-\alpha_n)$,we get:
$\frac{x^n+px+q}{x-\alpha_n} = (x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_{n-1})$.
Taking the limit as $x \to \alpha_n$ on both sides:
$\lim_{x \to \alpha_n} \frac{x^n+px+q}{x-\alpha_n} = (\alpha_n-\alpha_1)(\alpha_n-\alpha_2)\ldots(\alpha_n-\alpha_{n-1})$.
Using $L$'$H$ôpital's Rule on the left side:
$\lim_{x \to \alpha_n} \frac{\frac{d}{dx}(x^n+px+q)}{\frac{d}{dx}(x-\alpha_n)} = \lim_{x \to \alpha_n} (nx^{n-1}+p) = n\alpha_n^{n-1}+p$.
Thus,$(\alpha_n-\alpha_1)(\alpha_n-\alpha_2)\ldots(\alpha_n-\alpha_{n-1}) = n\alpha_n^{n-1}+p$.

(B) The given limit is $\lim _{x \rightarrow 1} \frac{\log x}{1-x}$.
Substituting $x = 1$,we get the indeterminate form $\frac{\log 1}{1-1} = \frac{0}{0}$.
Applying $L'\text{Hospital's Rule}$,we differentiate the numerator and the denominator with respect to $x$:
$\lim _{x \rightarrow 1} \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(1-x)} = \lim _{x \rightarrow 1} \frac{\frac{1}{x}}{-1}$.
Evaluating the limit as $x \rightarrow 1$:
$\frac{\frac{1}{1}}{-1} = -1$.

(C) $l_1 = \lim_{x \rightarrow 2^{+}} (x + [x]) = 2 + 2 = 4$.
$l_2 = \lim_{x \rightarrow 2^{-}} (2x - [x]) = 2(2) - 1 = 3$.
$l_3 = \lim_{x \rightarrow \pi/2} \frac{\cos x}{x - \pi/2}$. Using $L$'Hospital's rule,$\lim_{x \rightarrow \pi/2} \frac{-\sin x}{1} = -1$.
Thus,$l_3 < l_2 < l_1$.

(C) Given $f(x)=3 x^{15}-5 x^{10}+7 x^5+50 \cos (x-1)$.
We need to evaluate $L = \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{h^3+3 h}$.
Rewrite the limit as $L = \lim _{h \rightarrow 0} \left( \frac{f(1-h)-f(1)}{-h} \cdot \frac{-h}{h(h^2+3)} \right)$.
Since $\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} = f'(1)$,we have $L = f'(1) \cdot \lim _{h \rightarrow 0} \left( \frac{-1}{h^2+3} \right)$.
Now,$f'(x) = 45 x^{14} - 50 x^9 + 35 x^4 - 50 \sin (x-1)$.
Evaluating at $x=1$,$f'(1) = 45(1)^{14} - 50(1)^9 + 35(1)^4 - 50 \sin(0) = 45 - 50 + 35 = 30$.
Thus,$L = 30 \cdot \left( \frac{-1}{0^2+3} \right) = 30 \cdot \left( -\frac{1}{3} \right) = -10$.

(B) Given the limit is in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital$'s rule:
$\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3} = \lim _{x \rightarrow 9} \frac{\frac{d}{dx}(\sqrt{f(x)}-3)}{\frac{d}{dx}(\sqrt{x}-3)}$
$= \lim _{x \rightarrow 9} \frac{\frac{f^{\prime}(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}}$
$= \lim _{x \rightarrow 9} \frac{f^{\prime}(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$
Substituting the values $f(9)=9$ and $f^{\prime}(9)=4$:
$= \frac{f^{\prime}(9) \cdot \sqrt{9}}{\sqrt{f(9)}} = \frac{4 \cdot 3}{\sqrt{9}} = \frac{12}{3} = 4$.

(C) Given,$L = \lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3}$.
Checking the form at $x = 3$,we get $\frac{3^3 - 3(3^2) - 4(3) + 12}{2(3^3) - 7(3^2) + 2(3) + 3} = \frac{27 - 27 - 12 + 12}{54 - 63 + 6 + 3} = \frac{0}{0}$ form.
Applying $L'H\hat{o}pital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 3^{-}} \frac{\frac{d}{dx}(x^3-3 x^2-4 x+12)}{\frac{d}{dx}(2 x^3-7 x^2+2 x+3)} = \lim _{x \rightarrow 3^{-}} \frac{3 x^2-6 x-4}{6 x^2-14 x+2}$.
Now,substitute $x = 3$:
$L = \frac{3(3^2) - 6(3) - 4}{6(3^2) - 14(3) + 2} = \frac{27 - 18 - 4}{54 - 42 + 2} = \frac{5}{14}$.

(D) Given the limit: $\lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}$
Substituting $x = 2$,we get $\frac{2^3 - 2^2 - 2 - 2}{2(2^3) - 3(2^2) - 3(2) + 2} = \frac{8-4-2-2}{16-12-6+2} = \frac{0}{0}$ form.
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx}(x^3-x^2-x-2) = 3x^2-2x-1$
Denominator derivative: $\frac{d}{dx}(2x^3-3x^2-3x+2) = 6x^2-6x-3$
Now,evaluate the limit: $\lim _{x \rightarrow 2} \frac{3x^2-2x-1}{6x^2-6x-3}$
$= \frac{3(2)^2 - 2(2) - 1}{6(2)^2 - 6(2) - 3} = \frac{12-4-1}{24-12-3} = \frac{7}{9}$

(D) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{3^x(1 - \cos x) - (1 - \cos x)}$
$= \lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{(3^x - 1)(1 - \cos x)}$
$= \lim _{x \rightarrow 0} \frac{2^x - \sin x - 1}{(3^x - 1) \frac{(1 - \cos x)}{x^2}}$
Since $\lim _{x \rightarrow 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$,the expression becomes:
$= \lim _{x \rightarrow 0} \frac{2(2^x - \sin x - 1)}{3^x - 1}$
Applying $L$'Hospital's Rule:
$= \lim _{x \rightarrow 0} \frac{2(2^x \ln 2 - \cos x)}{3^x \ln 3}$
$= \frac{2(\ln 2 - 1)}{\ln 3}$
$= \frac{2}{\log 3}(\log 2 - 1)$

(A) Given: $f(3)=16$ and $f^{\prime}(3)=4$.
We need to evaluate the limit: $L = \lim _{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3}$.
Substituting $x=3$,we get the form $\frac{3f(3)-3f(3)}{3-3} = \frac{0}{0}$.
Applying $L^{\prime}$Hospital's Rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 3} \frac{\frac{d}{dx}(x f(3)-3 f(x))}{\frac{d}{dx}(x-3)}$.
$L = \lim _{x \rightarrow 3} \frac{f(3)-3 f^{\prime}(x)}{1}$.
Substituting $x=3$:
$L = f(3)-3 f^{\prime}(3) = 16 - 3(4) = 16 - 12 = 4$.

(A) Given $f(4)=4$ and $f^{\prime}(4)=16$.
Consider the limit $L = \lim _{x \rightarrow 4} \frac{\sqrt{f(x)}-2}{\sqrt{x}-2}$.
This is a $\left[\frac{0}{0}\right]$ indeterminate form.
Applying $L'\text{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 4} \frac{\frac{d}{dx}(\sqrt{f(x)}-2)}{\frac{d}{dx}(\sqrt{x}-2)} = \lim _{x \rightarrow 4} \frac{\frac{1}{2\sqrt{f(x)}} \cdot f^{\prime}(x)}{\frac{1}{2\sqrt{x}}}$.
Simplifying the expression:
$L = \lim _{x \rightarrow 4} \frac{f^{\prime}(x) \cdot \sqrt{x}}{\sqrt{f(x)}} = \frac{f^{\prime}(4) \cdot \sqrt{4}}{\sqrt{f(4)}}$.
Substituting the given values:
$L = \frac{16 \cdot 2}{\sqrt{4}} = \frac{32}{2} = 16$.

(B) For $\alpha = \lim_{x \rightarrow 0} \frac{x(2^x - 1)}{1 - \cos x}$. Applying $L'\text{Hôpital's rule}$:
$\alpha = \lim_{x \rightarrow 0} \frac{(2^x - 1) + x \cdot 2^x \ln 2}{\sin x}$.
Applying $L'\text{Hôpital's rule}$ again:
$\alpha = \lim_{x \rightarrow 0} \frac{2^x \ln 2 + 2^x \ln 2 + x \cdot 2^x (\ln 2)^2}{\cos x} = \frac{\ln 2 + \ln 2 + 0}{1} = 2 \ln 2$.
For $\beta = \lim_{x \rightarrow 0} \frac{x(2^x - 1)}{\sqrt{1+x^2} - \sqrt{1-x^2}}$. Rationalizing the denominator:
$\beta = \lim_{x \rightarrow 0} \frac{x(2^x - 1)(\sqrt{1+x^2} + \sqrt{1-x^2})}{(1+x^2) - (1-x^2)} = \lim_{x \rightarrow 0} \frac{x(2^x - 1)(\sqrt{1+x^2} + \sqrt{1-x^2})}{2x^2} = \lim_{x \rightarrow 0} \frac{(2^x - 1)}{x} \cdot \frac{(\sqrt{1+x^2} + \sqrt{1-x^2})}{2}$.
Since $\lim_{x \rightarrow 0} \frac{2^x - 1}{x} = \ln 2$,we have $\beta = \ln 2 \cdot \frac{1+1}{2} = \ln 2$.
Thus,$\alpha = 2\beta$.

(C) We evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$.
Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'Hospital's Rule:
$L = \lim _{x \rightarrow 0} \frac{\sec^2 x - \cos x}{3x^2}$.
This is still a $\frac{0}{0}$ form. Applying $L$'Hospital's Rule again:
$L = \lim _{x \rightarrow 0} \frac{2 \sec^2 x \tan x + \sin x}{6x}$.
Using the standard limits $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
$L = \lim _{x}$ ${\rightarrow 0} \left( \frac{2 \sec^2 x \cdot \tan x}{6x} + \frac{\sin x}{6x} \right) = \frac{2(1)(1)}{6} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

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

(A) Let $L = \lim _{x \rightarrow 0} \frac{2 \sin x - \sin 2x}{x^3}$ (which is of the form $\frac{0}{0}$).
Applying $L$-Hospital rule:
$L = \lim _{x \rightarrow 0} \frac{2 \cos x - 2 \cos 2x}{3x^2}$ (which is of the form $\frac{0}{0}$).
Applying $L$-Hospital rule again:
$L = \lim _{x \rightarrow 0} \frac{-2 \sin x + 4 \sin 2x}{6x}$ (which is of the form $\frac{0}{0}$).
Applying $L$-Hospital rule again:
$L = \lim _{x \rightarrow 0} \frac{-2 \cos x + 8 \cos 2x}{6}$.
Substituting $x = 0$:
$L = \frac{-2 \cos(0) + 8 \cos(0)}{6} = \frac{-2(1) + 8(1)}{6} = \frac{6}{6} = 1$.

(C) Let $f(x) = \sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right) + \sin k$.
As $x \rightarrow k$,the expression $[x] - [x/k]$ takes integer values.
Specifically,for $x$ in a small neighborhood of $k$,$[x] = k$ and $[x/k] = 1$ (since $k \geq 2$).
Thus,the term $2 \pi ([x] - [x/k])$ is an even multiple of $\pi$,say $2 \pi m$.
So,$\sin(2 \pi m - x) = \sin(-x) = -\sin x$.
The limit becomes $\lim_{x \rightarrow k} \frac{-\sin x + \sin k}{x - k}$.
Applying $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
$\lim_{x \rightarrow k} \frac{-\cos x}{1} = -\cos k$.

(C) Given the limit expression: $\lim _{x \rightarrow 0} \left[ \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x} \right] = k$
Using the property $\log(a^b) = b \log a$,we get:
$\lim _{x \rightarrow 0} \left[ \frac{(1+x) \log (1+x) - x}{x^2} \right] = k$
Since this is a $\frac{0}{0}$ form,we apply $L$' Hospital's Rule:
$\lim _{x \rightarrow 0} \frac{\frac{d}{dx} [(1+x) \log (1+x) - x]}{\frac{d}{dx} [x^2]} = k$
$\lim _{x \rightarrow 0} \frac{[1 \cdot \log (1+x) + (1+x) \cdot \frac{1}{1+x}] - 1}{2x} = k$
$\lim _{x \rightarrow 0} \frac{\log (1+x) + 1 - 1}{2x} = k$
$\lim _{x \rightarrow 0} \frac{\log (1+x)}{2x} = k$
$\frac{1}{2} \lim _{x \rightarrow 0} \frac{\log (1+x)}{x} = k$
Since $\lim _{x \rightarrow 0} \frac{\log (1+x)}{x} = 1$,we have $k = \frac{1}{2} \times 1 = \frac{1}{2}$.
Therefore,$12k = 12 \times \frac{1}{2} = 6$.

(B) Given the limit: $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$.
Applying $L^{\prime}$Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx}(a^x - x^a) = a^x \ln a - a x^{a-1}$.
Denominator derivative: $\frac{d}{dx}(x^x - a^a) = x^x(1 + \ln x)$.
Substituting $x = a$:
$\frac{a^a \ln a - a \cdot a^{a-1}}{a^a(1 + \ln a)} = -1$.
$\frac{a^a \ln a - a^a}{a^a(1 + \ln a)} = -1$.
Dividing numerator and denominator by $a^a$:
$\frac{\ln a - 1}{1 + \ln a} = -1$.
$\ln a - 1 = -(1 + \ln a)$.
$\ln a - 1 = -1 - \ln a$.
$2 \ln a = 0$.
$\ln a = 0 \Rightarrow a = e^0 = 1$.

(A) Let $L = \lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$.
Since the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$,we apply $L$-Hospital's rule:
$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(4^x-9^x)}{\frac{d}{dx}(x(4^x+9^x))}$
$L = \lim _{x \rightarrow 0} \frac{4^x \log 4 - 9^x \log 9}{(4^x+9^x) + x(4^x \log 4 + 9^x \log 9)}$
Substituting $x = 0$:
$L = \frac{4^0 \log 4 - 9^0 \log 9}{(4^0+9^0) + 0(4^0 \log 4 + 9^0 \log 9)}$
$L = \frac{\log 4 - \log 9}{1 + 1} = \frac{\log(4/9)}{2}$
$L = \frac{1}{2} \log(\frac{2^2}{3^2}) = \frac{1}{2} \cdot 2 \log(\frac{2}{3}) = \log(\frac{2}{3})$.

(B) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(10^x - 1)}{1 - \cos x}$.
Since this is a $\frac{0}{0}$ form,we apply $L$-Hospital's rule:
$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x \ln 10 + (10^x - 1)}{\sin x}$.
Again,this is a $\frac{0}{0}$ form,so we apply $L$-Hospital's rule again:
$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x (\ln 10)^2 + 10^x \ln 10 + 10^x \ln 10}{\cos x}$.
Substituting $x = 0$:
$L = \frac{0 \cdot 10^0 (\ln 10)^2 + 10^0 \ln 10 + 10^0 \ln 10}{\cos 0} = \frac{0 + \ln 10 + \ln 10}{1} = 2 \ln 10$.
Since $\log 10$ usually denotes $\log_{10} 10 = 1$ or $\ln 10$ depending on context,and given the options,the result is $2 \log 10$.

(B) We have $\lim _{x \rightarrow 0^{+}} x^{n} \ln x$.
This is an indeterminate form of type $0 \times \infty$.
We can rewrite it as $\lim _{x \rightarrow 0^{+}} \frac{\ln x}{x^{-n}}$.
Applying $L'\text{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:
$= \lim _{x \rightarrow 0^{+}} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x^{-n})} = \lim _{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-n x^{-n-1}}$.
$= \lim _{x \rightarrow 0^{+}} \frac{1}{x} \cdot \frac{x^{n+1}}{-n} = \lim _{x \rightarrow 0^{+}} \frac{x^{n}}{-n}$.
Since $n > 0$,as $x \rightarrow 0^{+}$,$x^{n} \rightarrow 0$.
Therefore,the limit is $\frac{0}{-n} = 0$.

(C) Let $L = \lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$.
This is an indeterminate form of type $\infty - \infty$.
Combining the terms,we get $L = \lim _{x \rightarrow 1} \frac{(x-1)-\ln x}{(x-1) \ln x}$.
This is of the form $\frac{0}{0}$. Applying $L'H\hat{o}pital's$ rule:
$L = \lim _{x \rightarrow 1} \frac{1 - \frac{1}{x}}{\ln x + (x-1) \cdot \frac{1}{x}} = \lim _{x \rightarrow 1} \frac{\frac{x-1}{x}}{\frac{x \ln x + x - 1}{x}} = \lim _{x \rightarrow 1} \frac{x-1}{x \ln x + x - 1}$.
Applying $L'H\hat{o}pital's$ rule again:
$L = \lim _{x \rightarrow 1} \frac{1}{\ln x + x \cdot \frac{1}{x} + 1} = \lim _{x \rightarrow 1} \frac{1}{\ln x + 1 + 1} = \frac{1}{0 + 2} = \frac{1}{2}$.

(C) Given the limit $L = \lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$.
Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule:
$L = \lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-6 f^{\prime}(2 x)+4 f^{\prime}(4 x)}{2 x}$
$L = \lim _{x \rightarrow 0} \frac{f^{\prime}(x)-3 f^{\prime}(2 x)+2 f^{\prime}(4 x)}{x}$
Applying $L$'$H$ôpital's rule again:
$L = \lim _{x}$ ${\rightarrow 0} \frac{f^{\prime \prime}(x)-3 f^{\prime \prime}(2 x) \cdot 2+2 f^{\prime \prime}(4 x) \cdot 4}{1}$
$L = f^{\prime \prime}(0)-6 f^{\prime \prime}(0)+8 f^{\prime \prime}(0)$
$L = k-6 k+8 k = 3 k$.

(B) Let $f(x) = \int_{2}^{x} 3 t^{2} dt$. The limit is of the form $\frac{0}{0}$ as $x \rightarrow 2$.
Using $L$' Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 2} \frac{\frac{d}{dx} \int_{2}^{x} 3 t^{2} dt}{\frac{d}{dx} (x-2)}$
By the Leibniz Integral Rule,$\frac{d}{dx} \int_{2}^{x} 3 t^{2} dt = 3 x^{2}$.
So,the limit becomes $\lim _{x \rightarrow 2} \frac{3 x^{2}}{1}$.
Substituting $x = 2$,we get $3 \times (2)^{2} = 3 \times 4 = 12$.

(A) Given $\lim _{x \rightarrow 0} \frac{a x e^{x}-b \log (1+x)}{x^{2}}=3$. Since the limit exists and the denominator approaches $0$,the numerator must also approach $0$ as $x \rightarrow 0$.
Applying $L$'Hospital's rule once: $\lim _{x \rightarrow 0} \frac{a e^{x} + a x e^{x} - \frac{b}{1+x}}{2x} = 3$.
For the limit to exist,the numerator must be $0$ at $x=0$: $a(1) + a(0) - b = 0$ $\Rightarrow a - b = 0$ $\Rightarrow a = b$.
Applying $L$'Hospital's rule again: $\lim _{x \rightarrow 0} \frac{a e^{x} + a e^{x} + a x e^{x} + \frac{b}{(1+x)^{2}}}{2} = 3$.
Substituting $x=0$: $\frac{a + a + 0 + b}{2} = 3 \Rightarrow 2a + b = 6$.
Since $a = b$,we have $2a + a = 6$ $\Rightarrow 3a = 6$ $\Rightarrow a = 2$.
Thus,$b = 2$.

(C) Given,$f(0)=0$ and $f'(0)=2$.
Using $L$'Hopital's rule for the limit $\lim _{x \rightarrow 0} \frac{\sum_{k=1}^{2015} f(kx)}{x}$:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx} [f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)]}{1}$
$= \lim _{x \rightarrow 0} [f'(x) + 2f'(2x) + 3f'(3x) + \ldots + 2015f'(2015x)]$
$= f'(0) + 2f'(0) + 3f'(0) + \ldots + 2015f'(0)$
$= f'(0) [1 + 2 + 3 + \ldots + 2015]$
$= 2 \times \frac{2015(2015+1)}{2}$
$= 2015 \times 2016$.

(D) Given,$f^{\prime}(4)=5$.
We need to evaluate the limit $L = \lim _{x \rightarrow 2} \frac{f(4)-f\left(x^{2}\right)}{x-2}$.
Since the limit is in the $\frac{0}{0}$ form,we apply $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $x$.
$L = \lim _{x \rightarrow 2} \frac{\frac{d}{dx}[f(4)-f(x^2)]}{\frac{d}{dx}[x-2]}$
$L = \lim _{x \rightarrow 2} \frac{0 - f^{\prime}(x^2) \cdot 2x}{1}$
Substituting $x=2$,we get:
$L = -f^{\prime}(2^2) \cdot 2(2) = -f^{\prime}(4) \cdot 4$
Since $f^{\prime}(4)=5$,we have $L = -(5) \cdot 4 = -20$.

(B) Let $L = \lim _{x \rightarrow 0} \left\{\frac{\sqrt{1+x}}{x} - \sqrt{1+\frac{1}{x^{2}}}\right\}$
$L = \lim _{x \rightarrow 0} \left\{\frac{\sqrt{1+x} - \sqrt{x^{2}+1}}{x}\right\}$
Since the limit is in the $\frac{0}{0}$ form,we apply $L$-Hospital's Rule:
$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\sqrt{1+x} - \sqrt{x^{2}+1})}{\frac{d}{dx}(x)}$
$L = \lim _{x \rightarrow 0} \frac{\frac{1}{2\sqrt{1+x}} - \frac{2x}{2\sqrt{x^{2}+1}}}{1}$
$L = \lim _{x \rightarrow 0} \left( \frac{1}{2\sqrt{1+x}} - \frac{x}{\sqrt{x^{2}+1}} \right)$
Substituting $x = 0$:
$L = \frac{1}{2\sqrt{1+0}} - \frac{0}{\sqrt{0^{2}+1}} = \frac{1}{2} - 0 = \frac{1}{2}$

(B) Given limit is $\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$.
This is a $\frac{0}{0}$ indeterminate form.
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\pi^{x}-1)}{\frac{d}{dx}(\sqrt{1+x}-1)}$
$= \lim _{x \rightarrow 0} \frac{\pi^{x} \log _{e} \pi}{\frac{1}{2 \sqrt{1+x}}}$
$= \lim _{x \rightarrow 0} 2 \sqrt{1+x} \cdot \pi^{x} \log _{e} \pi$
Substituting $x = 0$:
$= 2 \sqrt{1+0} \cdot \pi^{0} \log _{e} \pi$
$= 2 \cdot 1 \cdot 1 \cdot \log _{e} \pi$
$= 2 \log _{e} \pi = \log _{e} \pi^{2}$.

(D) Let $L = \lim_{x \rightarrow 5} \frac{x f(5)-5 f(x)}{x-5}$.
Since the limit is in the $\frac{0}{0}$ form,we apply $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim_{x \rightarrow 5} \frac{\frac{d}{dx}(x f(5)-5 f(x))}{\frac{d}{dx}(x-5)}$
$L = \lim_{x \rightarrow 5} \frac{f(5)-5 f'(x)}{1}$
Substituting $x=5$:
$L = f(5)-5 f'(5)$
Given $f(5)=7$ and $f'(5)=7$:
$L = 7 - 5(7) = 7 - 35 = -28$.

(C) Let $L = \lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$.
Taking the natural logarithm on both sides:
$\log L = \lim _{x \rightarrow 0^{+}} \frac{\log \left(e^{x}+x\right)}{x}$.
Since the form is $\frac{0}{0}$,we apply $L$'Hospital's rule:
$\log L = \lim _{x \rightarrow 0^{+}} \frac{\frac{d}{dx} \log \left(e^{x}+x\right)}{\frac{d}{dx} x} = \lim _{x \rightarrow 0^{+}} \frac{\frac{e^{x}+1}{e^{x}+x}}{1}$.
Evaluating the limit as $x \rightarrow 0^{+}$:
$\log L = \frac{e^{0}+1}{e^{0}+0} = \frac{1+1}{1+0} = 2$.
Therefore,$L = e^{2}$.

(B) Let $y = \lim _{x \rightarrow 0} (\sin x)^{2 \tan x}$.
Taking the natural logarithm on both sides:
$\ln y = \lim _{x \rightarrow 0} 2 \tan x \ln(\sin x)$.
This is an indeterminate form of type $0 \times (-\infty)$. We rewrite it as:
$\ln y = 2 \lim _{x \rightarrow 0} \frac{\ln(\sin x)}{\cot x}$.
Applying $L'H\hat{o}pital's$ rule:
$\ln y = 2 \lim _{x \rightarrow 0} \frac{\frac{\cos x}{\sin x}}{-\csc^2 x} = 2 \lim _{x \rightarrow 0} \frac{\cos x / \sin x}{-1 / \sin^2 x} = 2 \lim _{x \rightarrow 0} (-\cos x \sin x)$.
$\ln y = 2 \times (-1 \times 0) = 0$.
Since $\ln y = 0$,we have $y = e^0 = 1$.