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

(B) To evaluate the limit $\mathop {\lim }\limits_{x \to a} \frac{{\sqrt {3x - a} - \sqrt {x + a} }}{{x - a}}$,we rationalize the numerator:
$= \mathop {\lim }\limits_{x \to a} \frac{{\sqrt {3x - a} - \sqrt {x + a} }}{{x - a}} \times \frac{{\sqrt {3x - a} + \sqrt {x + a} }}{{\sqrt {3x - a} + \sqrt {x + a} }}$
$= \mathop {\lim }\limits_{x \to a} \frac{{(3x - a) - (x + a)}}{{(x - a)(\sqrt {3x - a} + \sqrt {x + a})}}$
$= \mathop {\lim }\limits_{x \to a} \frac{{2x - 2a}}{{(x - a)(\sqrt {3x - a} + \sqrt {x + a})}}$
$= \mathop {\lim }\limits_{x \to a} \frac{{2(x - a)}}{{(x - a)(\sqrt {3x - a} + \sqrt {x + a})}}$
$= \frac{2}{{\sqrt {3a - a} + \sqrt {a + a}}} = \frac{2}{{\sqrt {2a} + \sqrt {2a}}} = \frac{2}{{2\sqrt {2a}}} = \frac{1}{{\sqrt {2a}}}$
Alternatively,using $L'\text{Hospital's rule}$:
$= \mathop {\lim }\limits_{x \to a} \frac{{\frac{d}{{dx}}(\sqrt {3x - a} - \sqrt {x + a})}}{{\frac{d}{{dx}}(x - a)}}$
$= \mathop {\lim }\limits_{x \to a} \frac{{\frac{3}{{2\sqrt {3x - a} }} - \frac{1}{{2\sqrt {x + a} }}}}{1}$
$= \frac{3}{{2\sqrt {2a} }} - \frac{1}{{2\sqrt {2a} }} = \frac{2}{{2\sqrt {2a} }} = \frac{1}{{\sqrt {2a}}}$

(A) Using $L'Hospital's$ rule,we differentiate the numerator and the denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 1} \frac{{\log x}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{dx}(\log x)}}{{\frac{d}{dx}(x - 1)}}$
$= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{x}}}{1}$
$= \frac{1}{1} = 1$
Thus,the correct option is $A$.

(A) Using $L'\text{Hospital's rule}$ as the limit is of the form $\frac{0}{0}$:
$\mathop {\lim }\limits_{x \to 0} \frac{{\log \cos x}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}(\log \cos x)}}{{\frac{d}{{dx}}(x)}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\cos x}} \cdot (-\sin x)}}{1}$
$= \mathop {\lim }\limits_{x \to 0} (-\tan x)$
$= -\tan(0) = 0$

(B) Given the limit $\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}}$.
Since $f(9) = 9$,the expression takes the indeterminate form $\frac{0}{0}$ as $x \to 9$.
Applying $L'\text{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 9} \frac{\frac{d}{dx}(\sqrt{f(x)} - 3)}{\frac{d}{dx}(\sqrt{x} - 3)} = \mathop {\lim }\limits_{x \to 9} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}} = \mathop {\lim }\limits_{x \to 9} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$.
Substituting the values $f(9) = 9$ and $f'(9) = 4$:
$= \frac{f'(9) \cdot \sqrt{9}}{\sqrt{f(9)}} = \frac{4 \cdot 3}{\sqrt{9}} = \frac{12}{3} = 4$.

(B) Using $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{{{{(1 + x)}^{1/2}} - 1}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}({2^x} - 1)}}{{\frac{d}{{dx}}({{(1 + x)}^{1/2}} - 1)}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{{2^x}\ln 2}}{{\frac{1}{2}{{(1 + x)}^{ - 1/2}}}}$
Substituting $x = 0$:
$= \frac{{{2^0}\ln 2}}{{\frac{1}{2}{{(1 + 0)}^{ - 1/2}}}} = \frac{{1 \cdot \ln 2}}{{\frac{1}{2}}} = 2\ln 2 = \ln({2^2}) = \ln 4$.

(A) We know that $\mathop {\lim }\limits_{u \to 0} \frac{e^u - 1}{u} = 1$ and $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$.
$\mathop {\lim }\limits_{x \to 0} \frac{e^{\sin x} - 1}{x} = \mathop {\lim }\limits_{x \to 0} \left( \frac{e^{\sin x} - 1}{\sin x} \times \frac{\sin x}{x} \right)$
$= \left( \mathop {\lim }\limits_{\sin x \to 0} \frac{e^{\sin x} - 1}{\sin x} \right) \times \left( \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} \right)$
$= 1 \times 1 = 1$.
Alternatively,using $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 0} \frac{e^{\sin x} - 1}{x} = \mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(e^{\sin x} - 1)}{\frac{d}{dx}(x)}$
$= \mathop {\lim }\limits_{x \to 0} \frac{\cos x \cdot e^{\sin x}}{1} = \cos(0) \cdot e^{\sin(0)} = 1 \cdot e^0 = 1$.

(B) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x}$,we can use $L'Hospital's$ rule since it is a $\frac{0}{0}$ indeterminate form.
Differentiating the numerator and denominator with respect to $x$:
Numerator: $\frac{d}{dx}(\sqrt{1+\sin x} - \sqrt{1-\sin x}) = \frac{\cos x}{2\sqrt{1+\sin x}} - \frac{-\cos x}{2\sqrt{1-\sin x}} = \frac{\cos x}{2\sqrt{1+\sin x}} + \frac{\cos x}{2\sqrt{1-\sin x}}$
Denominator: $\frac{d}{dx}(x) = 1$
Applying the limit as $x \to 0$:
$\mathop {\lim }\limits_{x \to 0} \left( \frac{\cos x}{2\sqrt{1+\sin x}} + \frac{\cos x}{2\sqrt{1-\sin x}} \right) = \frac{\cos 0}{2\sqrt{1+\sin 0}} + \frac{\cos 0}{2\sqrt{1-\sin 0}}$
$= \frac{1}{2\sqrt{1}} + \frac{1}{2\sqrt{1}} = \frac{1}{2} + \frac{1}{2} = 1.$

(A) Let $L = \mathop {\lim }\limits_{\alpha \to \pi /4} \frac{{\sin \alpha - \cos \alpha }}{{\alpha - \pi /4}}$.
We can rewrite the numerator as:
$\sin \alpha - \cos \alpha = \sqrt{2} \left( \sin \alpha \cdot \frac{1}{\sqrt{2}} - \cos \alpha \cdot \frac{1}{\sqrt{2}} \right) = \sqrt{2} \left( \sin \alpha \cos \frac{\pi}{4} - \cos \alpha \sin \frac{\pi}{4} \right) = \sqrt{2} \sin \left( \alpha - \frac{\pi}{4} \right)$.
Substituting this into the limit:
$L = \mathop {\lim }\limits_{\alpha \to \pi /4} \frac{\sqrt{2} \sin \left( \alpha - \frac{\pi}{4} \right)}{\alpha - \frac{\pi}{4}}$.
Using the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,where $x = \alpha - \frac{\pi}{4}$:
$L = \sqrt{2} \times 1 = \sqrt{2}$.
Alternatively,using $L$-Hospital's rule:
$L = \mathop {\lim }\limits_{\alpha \to \pi /4} \frac{\frac{d}{d\alpha}(\sin \alpha - \cos \alpha)}{\frac{d}{d\alpha}(\alpha - \pi/4)} = \mathop {\lim }\limits_{\alpha \to \pi /4} \frac{\cos \alpha + \sin \alpha}{1} = \cos \frac{\pi}{4} + \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(A) Let $L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \tan x \log \sin x$.
This is an indeterminate form of type $\infty \times 0$.
We can rewrite the expression as $L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{\log \sin x}{\cot x}$.
Applying $L'\text{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:
$L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{\frac{d}{dx}(\log \sin x)}{\frac{d}{dx}(\cot x)} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{\frac{1}{\sin x} \cos x}{-\csc^2 x}$.
Simplifying the expression:
$L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{\cot x}{-\csc^2 x} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} (-\cos x \sin x)$.
As $x \to \frac{\pi }{2}$,$\cos x \to 0$ and $\sin x \to 1$.
Therefore,$L = -(0 \times 1) = 0$.

(A) We need to evaluate $\mathop {\lim }\limits_{\theta \to \pi /2} (\sec \theta - \tan \theta )$.
Substituting $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get:
$\mathop {\lim }\limits_{\theta \to \pi /2} \left( \frac{1 - \sin \theta}{\cos \theta} \right)$.
As $\theta \to \pi /2$,this is a $\frac{0}{0}$ indeterminate form.
Using the identity $1 - \sin \theta = \left( \cos \frac{\theta}{2} - \sin \frac{\theta}{2} \right)^2$ and $\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = (\cos \frac{\theta}{2} - \sin \frac{\theta}{2})(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})$:
$\mathop {\lim }\limits_{\theta \to \pi /2} \frac{(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})^2}{(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})} = \mathop {\lim }\limits_{\theta \to \pi /2} \frac{\cos \frac{\theta}{2} - \sin \frac{\theta}{2}}{\cos \frac{\theta}{2} + \sin \frac{\theta}{2}}$.
Substituting $\theta = \pi / 2$,we get $\frac{\cos(\pi/4) - \sin(\pi/4)}{\cos(\pi/4) + \sin(\pi/4)} = \frac{1/\sqrt{2} - 1/\sqrt{2}}{1/\sqrt{2} + 1/\sqrt{2}} = \frac{0}{\sqrt{2}} = 0$.

(C) Using $L-Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x - x}}{{3x - \sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}(\tan 2x - x)}}{{\frac{d}{{dx}}(3x - \sin x)}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{2\sec^2 2x - 1}}{{3 - \cos x}}$
Substituting $x = 0$:
$= \frac{{2\sec^2(0) - 1}}{{3 - \cos(0)}} = \frac{{2(1)^2 - 1}}{{3 - 1}} = \frac{{2 - 1}}{2} = \frac{1}{2}$.

(B) Using the trigonometric identity $\cos C - \cos D = 2 \sin \left( \frac{C+D}{2} \right) \sin \left( \frac{D-C}{2} \right)$,we have:
$\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2 \sin \left( \frac{ax+bx}{2} \right) \sin \left( \frac{bx-ax}{2} \right)}}{{{x^2}}}$
$= 2 \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin \left( \frac{a+b}{2}x \right)}{x} \right) \left( \frac{\sin \left( \frac{b-a}{2}x \right)}{x} \right)$
$= 2 \cdot \left( \frac{a+b}{2} \right) \cdot \left( \frac{b-a}{2} \right) = 2 \cdot \frac{b^2 - a^2}{4} = \frac{b^2 - a^2}{2}$
Alternatively,applying $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{-a \sin ax + b \sin bx}}{{2x}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{-a^2 \cos ax + b^2 \cos bx}}{2} = \frac{-a^2(1) + b^2(1)}{2} = \frac{b^2 - a^2}{2}$

(A) We need to evaluate $L = \mathop {\lim }\limits_{x \to 0} x \log (\sin x)$.
As $x \to 0^+$,$\sin x \to 0^+$,so $\log(\sin x) \to -\infty$.
This is an indeterminate form of type $0 \times (-\infty)$.
We can rewrite the expression as $L = \mathop {\lim }\limits_{x \to 0} \frac{\log(\sin x)}{1/x}$.
Applying $L$'$H$ôpital's rule:
$L = \mathop {\lim }\limits_{x \to 0} \frac{\frac{1}{\sin x} \cdot \cos x}{-1/x^2} = \mathop {\lim }\limits_{x \to 0} -x^2 \cot x$.
$L = \mathop {\lim }\limits_{x \to 0} -x \cdot (x \cot x) = \mathop {\lim }\limits_{x \to 0} -x \cdot \frac{x}{\tan x}$.
Since $\mathop {\lim }\limits_{x \to 0} \frac{x}{\tan x} = 1$,we have $L = 0 \cdot 1 = 0$.

(A) We use the Taylor series expansion for $\sin x$:
$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$
Substituting this into the limit expression:
$\mathop {\lim }\limits_{x \to 0} \frac{(x - \frac{x^3}{6} + \frac{x^5}{120} - \dots) - x + \frac{x^3}{6}}{x^5}$
$= \mathop {\lim }\limits_{x \to 0} \frac{\frac{x^5}{120} - \dots}{x^5}$
$= \frac{1}{120}$.

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} \right]$.
Combine the terms into a single fraction:
$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{x - \log (1 + x)}{{x^2}}} \right]$.
Since this is a $\frac{0}{0}$ indeterminate form,we apply $L'\text{Hospital's rule}$:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(x - \log (1 + x))}{\frac{d}{dx}(x^2)} = \mathop {\lim }\limits_{x \to 0} \frac{1 - \frac{1}{1+x}}{2x}$.
Simplify the expression:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{1+x-1}{1+x}}{2x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{2x(1+x)} = \mathop {\lim }\limits_{x \to 0} \frac{1}{2(1+x)}$.
Substitute $x = 0$:
$\frac{1}{2(1+0)} = \frac{1}{2}$.

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to \alpha } \frac{{\sin x - \sin \alpha }}{{x - \alpha }}$.
This is an indeterminate form of type $\frac{0}{0}$ as $x \to \alpha$.
Applying $L'Hospital's$ rule,we differentiate the numerator and the denominator with respect to $x$:
$\frac{d}{dx}(\sin x - \sin \alpha) = \cos x$
$\frac{d}{dx}(x - \alpha) = 1$
Thus,the limit becomes:
$\mathop {\lim }\limits_{x \to \alpha } \frac{\cos x}{1} = \cos \alpha $.

(C) Given limit: $L = \mathop {\lim }\limits_{x \to \pi /2} \frac{2x - \pi}{\cos x}$.
Since the form is $\frac{0}{0}$,we apply $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \mathop {\lim }\limits_{x \to \pi /2} \frac{\frac{d}{dx}(2x - \pi)}{\frac{d}{dx}(\cos x)}$
$L = \mathop {\lim }\limits_{x \to \pi /2} \frac{2}{-\sin x}$
Substituting $x = \pi /2$:
$L = \frac{2}{-\sin(\pi /2)} = \frac{2}{-1} = -2$.
Thus,the correct option is $C$.

(D) Using the Taylor series expansion for $\sin x$:
$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
Substituting this into the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots) - x}{x^3}$
$= \mathop {\lim }\limits_{x \to 0} \frac{-\frac{x^3}{6} + \frac{x^5}{120} - \dots}{x^3}$
$= \mathop {\lim }\limits_{x \to 0} (-\frac{1}{6} + \frac{x^2}{120} - \dots)$
$= -\frac{1}{6}$
Alternatively,applying $L$-Hospital's rule three times gives the same result.

(B) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x - \sin x}}{{{x^2}\sin x}}$.
Since the form is $\frac{0}{0}$ as $x \to 0$,we apply $L'\text{Hospital's rule}$.
First derivative of numerator: $\frac{d}{dx}(x\cos x - \sin x) = \cos x - x\sin x - \cos x = -x\sin x$.
First derivative of denominator: $\frac{d}{dx}(x^2\sin x) = 2x\sin x + x^2\cos x$.
So,the limit becomes $\mathop {\lim }\limits_{x \to 0} \frac{-x\sin x}{2x\sin x + x^2\cos x} = \mathop {\lim }\limits_{x \to 0} \frac{-\sin x}{2\sin x + x\cos x}$.
Applying $L'\text{Hospital's rule}$ again (still $\frac{0}{0}$ form):
Numerator derivative: $\frac{d}{dx}(-\sin x) = -\cos x$.
Denominator derivative: $\frac{d}{dx}(2\sin x + x\cos x) = 2\cos x + \cos x - x\sin x = 3\cos x - x\sin x$.
Evaluating the limit: $\mathop {\lim }\limits_{x \to 0} \frac{-\cos x}{3\cos x - x\sin x} = \frac{-\cos(0)}{3\cos(0) - 0\sin(0)} = \frac{-1}{3(1) - 0} = -\frac{1}{3}$.

(C) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x}$,we can use the rationalization method or $L$-Hospital's rule.
Method $1$: Rationalization
Multiply the numerator and denominator by the conjugate $\frac{{{{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}}}}{{{{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}}}}$:
$\mathop {\lim }\limits_{x \to 0} \frac{{(1 + x) - (1 - x)}}{{x({{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}})}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{x({{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}})}} = \mathop {\lim }\limits_{x \to 0} \frac{2}{{{{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}}}} = \frac{2}{{1 + 1}} = 1$.
Method $2$: $L$-Hospital's rule
Since the limit is in the form $\frac{0}{0}$,differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}({{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}})}}{{\frac{d}{{dx}}(x)}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{2}{{(1 + x)}^{ - 1/2}} - ( - \frac{1}{2}{{(1 - x)}^{ - 1/2}})}}{1} = \mathop {\lim }\limits_{x \to 0} \left( \frac{1}{{2\sqrt {1 + x} }} + \frac{1}{{2\sqrt {1 - x} }} \right) = \frac{1}{2} + \frac{1}{2} = 1$.

(B) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}}$.
Since the form is $\frac{0}{0}$,we apply $L$-Hospital's rule:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}(\sqrt {1 + x} - \sqrt {1 - x} )}}{{\frac{d}{{dx}}({{\sin }^{ - 1}}x)}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{2\sqrt {1 + x} }} - (\frac{{ - 1}}{{2\sqrt {1 - x} }})}}{{\frac{1}{{\sqrt {1 - {x^2}} }}}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{2\sqrt {1 + x} }} + \frac{1}{{2\sqrt {1 - x} }}}}{{\frac{1}{{\sqrt {1 - {x^2}} }}}}$
Substituting $x = 0$:
$L = \frac{{\frac{1}{2} + \frac{1}{2}}}{1} = \frac{1}{1} = 1$.

(B) Let $L = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}}$.
Since the form is $\frac{0}{0}$,we apply $L$-Hospital's rule:
$L = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\frac{d}{{dx}}(\sqrt 2 \cos x - 1)}}{{\frac{d}{{dx}}(\cot x - 1)}}$
$L = \mathop {\lim }\limits_{x \to \pi /4} \frac{{-\sqrt 2 \sin x}}{{-\csc^2 x}}$
$L = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sqrt 2 \sin x}}{{\csc^2 x}} = \mathop {\lim }\limits_{x \to \pi /4} \sqrt 2 \sin x \sin^2 x = \sqrt 2 \sin^3 x$
Substituting $x = \pi /4$:
$L = \sqrt 2 \left( \frac{1}{{\sqrt 2 }} \right)^3 = \sqrt 2 \cdot \frac{1}{{2\sqrt 2 }} = \frac{1}{2}$.

(C) Given the limit: $\mathop {\lim }\limits_{x \to a} \frac{{\cos x - \cos a}}{{\cot x - \cot a}}$
Applying $L$'$H$ôpital's Rule by differentiating the numerator and denominator with respect to $x$:
Numerator: $\frac{d}{dx}(\cos x - \cos a) = -\sin x$
Denominator: $\frac{d}{dx}(\cot x - \cot a) = -\csc^2 x$
Thus,the limit becomes: $\mathop {\lim }\limits_{x \to a} \frac{-\sin x}{-\csc^2 x} = \mathop {\lim }\limits_{x \to a} \frac{\sin x}{\frac{1}{\sin^2 x}} = \mathop {\lim }\limits_{x \to a} \sin^3 x$
Evaluating the limit as $x \to a$: $\sin^3 a$.

(B) Let $y = x^x$. Taking the natural logarithm on both sides,we get $\log y = x \log x$.
Now,we evaluate the limit as $x \to 0^+$:
$\mathop {\lim }\limits_{x \to 0^+} \log y = \mathop {\lim }\limits_{x \to 0^+} x \log x = \mathop {\lim }\limits_{x \to 0^+} \frac{\log x}{1/x}$.
Using $L$'$H$ôpital's rule,we differentiate the numerator and denominator:
$= \mathop {\lim }\limits_{x \to 0^+} \frac{1/x}{-1/x^2} = \mathop {\lim }\limits_{x \to 0^+} (-x) = 0$.
Since $\log y \to 0$,it follows that $y \to e^0 = 1$.
Therefore,$\mathop {\lim }\limits_{x \to 0^+} x^x = 1$.

(A) We need to evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}}$.
Using the Taylor series expansions for ${e^{{x^2}}}$ and $\cos x$ around $x = 0$:
${e^{{x^2}}} = 1 + {x^2} + \frac{{{x^4}}}{{2!}} + \dots$
$\cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - \dots$
Substituting these into the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{{(1 + {x^2} + \frac{{{x^4}}}{2} + \dots) - (1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} - \dots)}}{{{x^2}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{{x^2} + \frac{{{x^2}}}{2} + O({x^4})}}{{{x^2}}}$
$= \mathop {\lim }\limits_{x \to 0} (1 + \frac{1}{2} + O({x^2})) = \frac{3}{2}$.
Alternatively,using $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x{e^{{x^2}}} + \sin x}}{{2x}}$
$= \mathop {\lim }\limits_{x \to 0} \left( {e^{{x^2}}} + \frac{{\sin x}}{{2x}} \right)$
$= 1 + \frac{1}{2} = \frac{3}{2}$.

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to a} \frac{x^{-1} - a^{-1}}{x - a}$.
Step $1$: Simplify the expression inside the limit.
$\frac{x^{-1} - a^{-1}}{x - a} = \frac{\frac{1}{x} - \frac{1}{a}}{x - a} = \frac{\frac{a - x}{ax}}{x - a}$.
Step $2$: Simplify the fraction.
$\frac{-(x - a)}{ax(x - a)} = -\frac{1}{ax}$ (for $x \neq a$).
Step $3$: Apply the limit as $x \to a$.
$\mathop {\lim }\limits_{x \to a} (-\frac{1}{ax}) = -\frac{1}{a(a)} = -\frac{1}{a^2}$.
Alternatively,using $L'\text{Hospital's rule}$:
$\mathop {\lim }\limits_{x \to a} \frac{\frac{d}{dx}(x^{-1} - a^{-1})}{\frac{d}{dx}(x - a)} = \mathop {\lim }\limits_{x \to a} \frac{-x^{-2}}{1} = -\frac{1}{a^2}$.

(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (1 - \sin x)\tan x$.
Rewrite the expression as: $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (1 - \sin x) \frac{\sin x}{\cos x} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{\sin x - \sin^2 x}{\cos x}$.
As $x \to \frac{\pi }{2}$,the expression takes the form $\frac{1-1}{0} = \frac{0}{0}$,which is an indeterminate form.
Applying $L'\text{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{\frac{d}{dx}(\sin x - \sin^2 x)}{\frac{d}{dx}(\cos x)} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{\cos x - 2\sin x \cos x}{-\sin x}$.
Substituting $x = \frac{\pi }{2}$:
$\frac{\cos(\frac{\pi }{2}) - 2\sin(\frac{\pi }{2})\cos(\frac{\pi }{2})}{-\sin(\frac{\pi }{2})} = \frac{0 - 2(1)(0)}{-1} = \frac{0}{-1} = 0$.

(C) Using $L$-Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x + \log (1 - x)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x - \frac{1}{{1 - x}}}}{{2x}}$
Applying $L$-Hospital's rule again:
$= \mathop {\lim }\limits_{x \to 0} \frac{{ - \sin x - \frac{1}{{{{(1 - x)}^2}}}}}{2} = \frac{{ - 0 - 1}}{2} = - \frac{1}{2}$.
Alternatively,using Taylor series expansions:
$\sin x = x - \frac{{{x^3}}}{{3!}} + \dots$
$\log (1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \dots$
Substituting these into the limit:
$= \mathop {\lim }\limits_{x \to 0} \frac{{(x - \frac{{{x^3}}}{6} + \dots) + (- x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \dots)}}{{{x^2}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{ - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{2} - \dots}}{{{x^2}}} = - \frac{1}{2}$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} {\left\{ {\tan \left( {\frac{\pi }{4} + x} \right)} \right\}^{1/x}}$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get $\tan(\frac{\pi}{4} + x) = \frac{1 + \tan x}{1 - \tan x}$.
So,$L = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + \tan x}}{{1 - \tan x}}} \right)^{1/x}}$.
This is of the form $1^\infty$,so $L = e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x} (\frac{1 + \tan x}{1 - \tan x} - 1)}$.
$L = e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x} (\frac{1 + \tan x - 1 + \tan x}{1 - \tan x})} = e^{\mathop {\lim }\limits_{x \to 0} \frac{2 \tan x}{x(1 - \tan x)}}$.
Since $\mathop {\lim }\limits_{x \to 0} \frac{\tan x}{x} = 1$,we have $L = e^{2 \times 1 \times \frac{1}{1 - 0}} = e^2$.

(A) We are given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}}$.
Rewrite the numerator by factoring out $e^x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x}({e^{\tan x - x}} - 1)}}{{\tan x - x}}$.
Using the property of limits $\mathop {\lim }\limits_{u \to 0} \frac{e^u - 1}{u} = 1$,where $u = \tan x - x$:
As $x \to 0$,$u = \tan x - x \to 0$.
Thus,the expression becomes:
$\mathop {\lim }\limits_{x \to 0} {e^x} \times \mathop {\lim }\limits_{u \to 0} \frac{e^u - 1}{u} = {e^0} \times 1 = 1 \times 1 = 1$.

(D) Let $y = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x} + {c^x}}}{3}} \right)^{2/x}}$.
Taking the natural logarithm on both sides:
$\ln y = \mathop {\lim }\limits_{x \to 0} \frac{2}{x} \ln \left( \frac{a^x + b^x + c^x}{3} \right)$.
Since this is a $\frac{0}{0}$ form,we apply $L$'$H$ôpital's rule:
$\ln y = 2 \mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx} \ln \left( \frac{a^x + b^x + c^x}{3} \right)}{\frac{d}{dx} (x)}$.
$\ln y = 2 \mathop {\lim }\limits_{x \to 0} \frac{3}{a^x + b^x + c^x} \cdot \frac{a^x \ln a + b^x \ln b + c^x \ln c}{3}$.
As $x \to 0$,$a^x, b^x, c^x \to 1$:
$\ln y = 2 \cdot \frac{1}{3} (\ln a + \ln b + \ln c) = \frac{2}{3} \ln(abc) = \ln((abc)^{2/3})$.
Therefore,$y = (abc)^{2/3}$.

(A) We evaluate the limit $\mathop {\lim }\limits_{x \to 0^ + } x^m (\log x)^n$ as $x \to 0^+$.
This is an indeterminate form of type $0 \times \infty$.
We rewrite the expression as $\mathop {\lim }\limits_{x \to 0^ + } \frac{(\log x)^n}{x^{-m}}$,which is of the form $\frac{\infty}{\infty}$.
Applying $L'\text{Hospital's rule}$ $n$ times,we differentiate the numerator and denominator with respect to $x$ repeatedly.
The $k$-th derivative of $(\log x)^n$ involves terms of $(\log x)^{n-k}$.
After $n$ applications of $L'\text{Hospital's rule}$,the numerator becomes a constant $n!$ and the denominator remains a power of $x$ (specifically $x^{-m}$ multiplied by constants).
Specifically,$\mathop {\lim }\limits_{x \to 0^ + } \frac{n! \cdot (-1)^n}{(-m)^n x^{-m}} = \mathop {\lim }\limits_{x \to 0^ + } \frac{n! \cdot (-1)^n x^m}{(-m)^n}$.
Since $m > 0$,as $x \to 0^+$,$x^m \to 0$.
Therefore,the limit is $0$.

(A) Given the limit: $\mathop {\lim }\limits_{x \to \infty } \frac{\log x}{x^n}$ where $n > 0$.
Since the form is $\frac{\infty}{\infty}$ as $x \to \infty$,we apply $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to \infty } \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(x^n)} = \mathop {\lim }\limits_{x \to \infty } \frac{\frac{1}{x}}{n x^{n-1}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{1}{n x^n}$
As $x \to \infty$,$x^n \to \infty$ for $n > 0$,so the expression approaches $0$.

(A) We are given the limit $L = \mathop {\lim }\limits_{x \to a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$.
Since the expression is in the form $\frac{-\infty}{-\infty}$ as $x \to a^+$,we apply $L$'$H$ôpital's Rule:
$L = \mathop {\lim }\limits_{x \to a} \frac{\frac{d}{dx}[\log(x-a)]}{\frac{d}{dx}[\log(e^x - e^a)]} = \mathop {\lim }\limits_{x \to a} \frac{\frac{1}{x-a}}{\frac{e^x}{e^x - e^a}} = \mathop {\lim }\limits_{x \to a} \frac{e^x - e^a}{(x-a)e^x}$.
This is now in the $\frac{0}{0}$ form. Applying $L$'$H$ôpital's Rule again:
$L = \mathop {\lim }\limits_{x \to a} \frac{e^x}{e^x + (x-a)e^x} = \frac{e^a}{e^a + (a-a)e^a} = \frac{e^a}{e^a} = 1$.

(B) Let $L = \mathop {\lim }\limits_{x \to \pi /2} \left[ {x\tan x - \left( {\frac{\pi }{2}} \right)\sec x} \right]$
$L = \mathop {\lim }\limits_{x \to \pi /2} \left[ {x \frac{\sin x}{\cos x} - \frac{\pi}{2} \frac{1}{\cos x}} \right] = \mathop {\lim }\limits_{x \to \pi /2} \frac{2x \sin x - \pi}{2 \cos x}$
This is in the indeterminate form $\frac{0}{0}$ as $x \to \pi/2$.
Applying $L$-Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \mathop {\lim }\limits_{x \to \pi /2} \frac{\frac{d}{dx}(2x \sin x - \pi)}{\frac{d}{dx}(2 \cos x)} = \mathop {\lim }\limits_{x \to \pi /2} \frac{2 \sin x + 2x \cos x}{-2 \sin x}$
Substituting $x = \pi/2$:
$L = \frac{2 \sin(\pi/2) + 2(\pi/2) \cos(\pi/2)}{-2 \sin(\pi/2)} = \frac{2(1) + \pi(0)}{-2(1)} = \frac{2}{-2} = -1$

(D) Let $y = \mathop {\lim }\limits_{x \to 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$.
Since this is a $\frac{0}{0}$ form,we apply $L'\text{Hospital's rule}$.
$y = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}(x{e^x} - \log (1 + x))}}{{\frac{d}{{dx}}({x^2})}}$
$y = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + x{e^x} - \frac{1}{{1 + x}}}}{{2x}}$.
Again,this is a $\frac{0}{0}$ form,so we apply $L'\text{Hospital's rule}$ again.
$y = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}({e^x} + x{e^x} - \frac{1}{{1 + x}})}}{{\frac{d}{{dx}}(2x)}}$
$y = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^x} + x{e^x} + \frac{1}{{{{(1 + x)}^2}}}}}{2}$
Substituting $x = 0$:
$y = \frac{{{e^0} + {e^0} + 0 \cdot {e^0} + \frac{1}{{{{(1 + 0)}^2}}}}}{2} = \frac{{1 + 1 + 0 + 1}}{2} = \frac{3}{2}$.

(D) Let $y = \lim_{x \to 3} \frac{x^3 - x^2 - 18}{x - 3}$.
Substituting $x = 3$,we get $\frac{3^3 - 3^2 - 18}{3 - 3} = \frac{27 - 9 - 18}{0} = \frac{0}{0}$ form.
Applying $L'\text{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:
$y = \lim_{x \to 3} \frac{\frac{d}{dx}(x^3 - x^2 - 18)}{\frac{d}{dx}(x - 3)}$
$y = \lim_{x \to 3} \frac{3x^2 - 2x}{1}$
Substituting $x = 3$:
$y = 3(3)^2 - 2(3) = 3(9) - 6 = 27 - 6 = 21$.

(D) We are given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x}$.
Substituting $x = 0$,we get the indeterminate form $\frac{0}{0}$.
Using $L$'$H$ôpital's rule,we differentiate the numerator and the denominator with respect to $x$:
$\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}$ and $\frac{d}{dx}(x) = 1$.
Thus,$\mathop {\lim }\limits_{x \to 0} \frac{\frac{1}{1+x^2}}{1} = \mathop {\lim }\limits_{x \to 0} \frac{1}{1+x^2}$.
Evaluating the limit as $x \to 0$,we get $\frac{1}{1+0^2} = 1$.

(A) Given limit: $\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x - \log (1 + x)}}{{{x^2}}}$,which is of the form $\frac{0}{0}$.
Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{dx}(x\cos x - \log (1 + x))}}{{\frac{d}{dx}(x^2)}} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x - x\sin x - \frac{1}{{1 + x}}}}{{2x}}$.
This is still of the form $\frac{0}{0}$. Applying $L'Hospital's$ rule again:
$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{dx}(\cos x - x\sin x - \frac{1}{{1 + x}})}}{{\frac{d}{dx}(2x)}} = \mathop {\lim }\limits_{x \to 0} \frac{{-\sin x - (\sin x + x\cos x) + \frac{1}{{{{(1 + x)}^2}}}}}{2}$.
Evaluating the limit as $x \to 0$:
$= \frac{{-\sin(0) - (\sin(0) + 0 \cdot \cos(0)) + \frac{1}{{{{(1 + 0)}^2}}}}}{2} = \frac{{0 - 0 + 1}}{2} = \frac{1}{2}$.

(D) The given limit is of the form $\frac{0}{0}$ as $x \to 1$.
Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 1} \frac{\frac{d}{dx}(1 + \log x - x)}{\frac{d}{dx}(1 - 2x + x^2)} = \mathop {\lim }\limits_{x \to 1} \frac{\frac{1}{x} - 1}{-2 + 2x}$
$= \mathop {\lim }\limits_{x \to 1} \frac{1 - x}{x(-2 + 2x)} = \mathop {\lim }\limits_{x \to 1} \frac{-(x - 1)}{2x(x - 1)}$
$= \mathop {\lim }\limits_{x \to 1} \frac{-1}{2x} = -\frac{1}{2}$.

(D) Given limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x}}{{{x^3}}}$,which is of the form $\left( \frac{0}{0} \right)$.
Applying $L'\text{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{1 + {x^2}}}}}{{3{x^2}}}$,which is still of the form $\left( \frac{0}{0} \right)$.
Applying $L'\text{Hospital's rule}$ again:
$= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}} \left( (1 - {x^2})^{-1/2} - (1 + {x^2})^{-1} \right)}}{{6x}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{-\frac{1}{2}(1 - {x^2})^{-3/2}(-2x) - (-1)(1 + {x^2})^{-2}(2x)}}{{6x}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{x(1 - {x^2})^{-3/2} + 2x(1 + {x^2})^{-2}}}{{6x}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{1}{6} \left[ (1 - {x^2})^{-3/2} + 2(1 + {x^2})^{-2} \right]$
$= \frac{1}{6} [1 + 2] = \frac{3}{6} = \frac{1}{2}$.

(D) The given limit is of the form $\frac{0}{0}$ as $x \to 0$.
Applying $L'Hospital's$ rule,we differentiate the numerator and the denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\cos x}} \cdot (-\sin x)}}{{2x}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{-\tan x}}{{2x}}$
Applying $L'Hospital's$ rule again:
$= \mathop {\lim }\limits_{x \to 0} \frac{{-\sec^2 x}}{2}$
$= \frac{{-1^2}}{2} = -\frac{1}{2}$.

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{\alpha \to \beta } \frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}$
Since the limit is in the indeterminate form $\frac{0}{0}$ as $\alpha \to \beta$,we apply $L'\text{Hospital's rule}$ by differentiating the numerator and the denominator with respect to $\alpha$:
$\frac{d}{d\alpha} (\sin^2 \alpha - \sin^2 \beta) = 2 \sin \alpha \cos \alpha = \sin 2\alpha$
$\frac{d}{d\alpha} (\alpha^2 - \beta^2) = 2\alpha$
Now,taking the limit as $\alpha \to \beta$:
$\mathop {\lim }\limits_{\alpha \to \beta } \frac{{\sin 2\alpha }}{{2\alpha }} = \frac{{\sin 2\beta }}{{2\beta }}$.

(B) Let $L = \mathop {\lim }\limits_{x \to 1} \frac{{1 + \cos \pi x}}{{{{\tan }^2}\pi x}}$.
This is a $\frac{0}{0}$ form.
Using $L$-Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
$L = \mathop {\lim }\limits_{x \to 1} \frac{{-\pi \sin \pi x}}{{2 \tan \pi x \cdot \sec^2 \pi x \cdot \pi}}$
$L = \mathop {\lim }\limits_{x \to 1} \frac{{-\sin \pi x}}{{2 \tan \pi x \cdot \sec^2 \pi x}}$
Since $\tan \pi x = \frac{\sin \pi x}{\cos \pi x}$,we have:
$L = \mathop {\lim }\limits_{x \to 1} \frac{{-\sin \pi x \cdot \cos \pi x}}{{2 \sin \pi x \cdot \sec^2 \pi x}}$
$L = \mathop {\lim }\limits_{x \to 1} \frac{{-\cos \pi x}}{{2 \sec^2 \pi x}} = \mathop {\lim }\limits_{x \to 1} \frac{{-\cos^3 \pi x}}{2}$
Substituting $x = 1$:
$L = \frac{{-\cos^3 \pi}}{2} = \frac{{-(-1)^3}}{2} = \frac{1}{2}$.

(D) Given limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\log _e}(1 + x)}{{3^x - 1}}$
This is in the $\frac{0}{0}$ indeterminate form.
Applying $L'\text{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}[\log _e(1 + x)]}{\frac{d}{dx}[3^x - 1]} = \mathop {\lim }\limits_{x \to 0} \frac{\frac{1}{1 + x}}{3^x \log _e 3}$
Substituting $x = 0$:
$= \frac{\frac{1}{1 + 0}}{3^0 \log _e 3} = \frac{1}{1 \cdot \log _e 3} = \frac{1}{\log _e 3}$
Using the property $\frac{1}{\log _a b} = \log _b a$,we get:
$= \log _3 e$.

(C) Let $L = \mathop {\lim }\limits_{x \to 2} \frac{{xf(2) - 2f(x)}}{{x - 2}}$.
Since the limit is of the form $\frac{0}{0}$ at $x = 2$,we can apply $L$'$H$ôpital's rule or manipulate the expression.
Adding and subtracting $2f(2)$ in the numerator:
$L = \mathop {\lim }\limits_{x \to 2} \frac{{xf(2) - 2f(2) + 2f(2) - 2f(x)}}{{x - 2}}$
$L = \mathop {\lim }\limits_{x \to 2} \left[ \frac{f(2)(x - 2)}{x - 2} - 2\frac{f(x) - f(2)}{x - 2} \right]$
$L = f(2) - 2 \mathop {\lim }\limits_{x \to 2} \frac{f(x) - f(2)}{x - 2}$
$L = f(2) - 2f'(2)$
Given $f(2) = 4$ and $f'(2) = 4$:
$L = 4 - 2(4) = 4 - 8 = -4$.

(A) We are given the limit $\mathop {\lim }\limits_{x \to \infty } \frac{{\log {x^n} - [x]}}{{[x]}}.$
This can be rewritten as $\mathop {\lim }\limits_{x \to \infty } \left( \frac{{\log {x^n}}}{{[x]}} - \frac{{[x]}}{{[x]}} \right).$
Since $x - 1 < [x] \le x,$ as $x \to \infty,$ $[x] \approx x.$
Thus,$\mathop {\lim }\limits_{x \to \infty } \frac{{\log {x^n}}}{{[x]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n \log x}}{x}.$
Using $L$'Hopital's rule or the standard limit $\mathop {\lim }\limits_{x \to \infty } \frac{{\log x}}{x} = 0,$ we get $n \times 0 = 0.$
Therefore,the expression becomes $0 - 1 = -1.$

(A) Let $y = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$.
Multiplying the numerator and denominator by the conjugates $(\sqrt{f(x)} + 1)$ and $(\sqrt{x} + 1)$:
$y = \mathop {\lim }\limits_{x \to 1} \frac{(\sqrt{f(x)} - 1)(\sqrt{f(x)} + 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)} \times \frac{\sqrt{x} + 1}{\sqrt{f(x)} + 1}$
$y = \mathop {\lim }\limits_{x \to 1} \frac{f(x) - 1}{x - 1} \times \frac{\sqrt{x} + 1}{\sqrt{f(x)} + 1}$
Since $f(1) = 1$,we can write $f(x) - 1$ as $f(x) - f(1)$:
$y = \left( \mathop {\lim }\limits_{x \to 1} \frac{f(x) - f(1)}{x - 1} \right) \times \left( \mathop {\lim }\limits_{x \to 1} \frac{\sqrt{x} + 1}{\sqrt{f(x)} + 1} \right)$
$y = f'(1) \times \frac{\sqrt{1} + 1}{\sqrt{f(1)} + 1} = 2 \times \frac{2}{1 + 1} = 2 \times 1 = 2$.
Alternatively,using $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 1} \frac{\frac{1}{2\sqrt{f(x)}} f'(x)}{\frac{1}{2\sqrt{x}}} = \mathop {\lim }\limits_{x \to 1} \frac{f'(x) \sqrt{x}}{\sqrt{f(x)}} = \frac{f'(1) \sqrt{1}}{\sqrt{f(1)}} = \frac{2 \times 1}{1} = 2$.

(A) Let $y = \mathop {\lim }\limits_{x \to 0} \frac{{{4^x} - {9^x}}}{{x({4^x} + {9^x})}}$. This is a $\frac{0}{0}$ indeterminate form.
Applying $L'Hospital's$ rule by differentiating the numerator and denominator with respect to $x$:
$y = \mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(4^x - 9^x)}{\frac{d}{dx}(x(4^x + 9^x))}$
$y = \mathop {\lim }\limits_{x \to 0} \frac{4^x \ln 4 - 9^x \ln 9}{(4^x + 9^x) + x(4^x \ln 4 + 9^x \ln 9)}$
Substituting $x = 0$:
$y = \frac{4^0 \ln 4 - 9^0 \ln 9}{(4^0 + 9^0) + 0(4^0 \ln 4 + 9^0 \ln 9)}$
$y = \frac{\ln 4 - \ln 9}{1 + 1} = \frac{\ln(4/9)}{2}$
$y = \frac{\ln((2/3)^2)}{2} = \frac{2 \ln(2/3)}{2} = \ln \left( \frac{2}{3} \right)$.

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}}$.
Method $1$: Using standard limits.
$\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = \mathop {\lim }\limits_{x \to 0} \left( \frac{\frac{a^x - 1}{x} - \frac{b^x - 1}{x}}{\frac{e^x - 1}{x}} \right)$.
Using the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{k^x - 1}{x} = \ln k$,we get:
$= \frac{\ln a - \ln b}{\ln e} = \ln \left( \frac{a}{b} \right)$.
Method $2$: Using $L-Hospital's$ rule.
Since the form is $\frac{0}{0}$,differentiate numerator and denominator with respect to $x$:
$= \mathop {\lim }\limits_{x \to 0} \frac{a^x \ln a - b^x \ln b}{e^x} = \frac{a^0 \ln a - b^0 \ln b}{e^0} = \ln a - \ln b = \ln \left( \frac{a}{b} \right)$.