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

(C) Let $y = \mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$.
Since the limit is in the $\frac{0}{0}$ form,we apply $L'\text{Hospital's rule}$.
$y = \mathop {\lim }\limits_{x \to - 2} \frac{\frac{d}{dx}(\sin^{-1}(x+2))}{\frac{d}{dx}(x^2+2x)}$
$y = \mathop {\lim }\limits_{x \to - 2} \frac{\frac{1}{\sqrt{1-(x+2)^2}}}{2x+2}$
Substituting $x = -2$:
$y = \frac{\frac{1}{\sqrt{1-0}}}{2(-2)+2} = \frac{1}{-4+2} = -\frac{1}{2}$.

(D) Applying $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}$
$= \mathop {\lim }\limits_{x \to 7} \frac{{\frac{d}{{dx}}(2 - \sqrt {x - 3} )}}{{\frac{d}{{dx}}({x^2} - 49)}}$
$= \mathop {\lim }\limits_{x \to 7} \frac{{0 - \frac{1}{{2\sqrt {x - 3} }}}}{{2x}}$
$= \mathop {\lim }\limits_{x \to 7} \frac{{ - 1}}{{4x\sqrt {x - 3} }}$
$= \frac{{ - 1}}{{4(7)\sqrt {7 - 3} }} = \frac{{ - 1}}{{28\sqrt 4 }} = \frac{{ - 1}}{{28(2)}} = -\frac{1}{56}$.

(C) Given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\log (3 + x) - \log (3 - x)}}{x} = k$.
Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(\log(3+x) - \log(3-x))}{\frac{d}{dx}(x)} = k$.
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{1}{3+x} - (\frac{1}{3-x} \times -1)}{1} = k$.
$\mathop {\lim }\limits_{x \to 0} (\frac{1}{3+x} + \frac{1}{3-x}) = k$.
Substituting $x = 0$:
$\frac{1}{3+0} + \frac{1}{3-0} = k$.
$\frac{1}{3} + \frac{1}{3} = k$.
$k = \frac{2}{3}$.

(D) Using $L$'Hopital's rule,we differentiate the numerator and denominator with respect to $h$:
$\mathop {\lim }\limits_{h \to 0} \frac{{f(2h + 2 + {h^2}) - f(2)}}{{f(h - {h^2} + 1) - f(1)}} = \mathop {\lim }\limits_{h \to 0} \frac{{f'(2h + 2 + {h^2}) \cdot (2 + 2h)}}{{f'(h - {h^2} + 1) \cdot (1 - 2h)}}$
As $h \to 0$,the expression becomes:
$\frac{{f'(2) \cdot 2}}{{f'(1) \cdot 1}} = \frac{{6 \times 2}}{{4 \times 1}} = \frac{{12}}{{4}} = 3$.

(C) Let $y = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$.
Using the series expansion for $e^x$ and $e^{-x}$:
$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$
Subtracting the two series:
$e^x - e^{-x} = 2 \left( x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots \right)$
Substituting this into the limit:
$y = \mathop {\lim }\limits_{x \to 0} \frac{2 \left( x + \frac{x^3}{6} + \dots \right)}{\sin x}$
Divide numerator and denominator by $x$:
$y = \mathop {\lim }\limits_{x \to 0} \frac{2 \left( 1 + \frac{x^2}{6} + \dots \right)}{\frac{\sin x}{x}}$
Since $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$:
$y = \frac{2(1 + 0)}{1} = 2$.
Alternatively,using $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}({e^x} - {e^{ - x}})}{\frac{d}{dx}(\sin x)}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{{\cos x}} = \frac{{{e^0} + {e^0}}}{{\cos 0}} = \frac{1 + 1}{1} = 2$.

(B) Given the limit: $\mathop {\lim }\limits_{x \to \pi /6} \frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}$.
Since the form is $0/0$,we apply $L-Hospital's$ rule by differentiating the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx}(3\sin x - \sqrt 3 \cos x) = 3\cos x + \sqrt 3 \sin x$.
Denominator derivative: $\frac{d}{dx}(6x - \pi) = 6$.
Applying the limit as $x \to \pi /6$:
$\frac{3\cos(\pi /6) + \sqrt 3 \sin(\pi /6)}{6} = \frac{3(\sqrt 3 /2) + \sqrt 3 (1/2)}{6} = \frac{(3\sqrt 3 + \sqrt 3)/2}{6} = \frac{4\sqrt 3}{12} = \frac{\sqrt 3}{3} = \frac{1}{\sqrt 3}$.

(C) To evaluate the limit $\mathop {\lim }\limits_{\theta \to \frac{\pi }{2}} \frac{\frac{\pi }{2} - \theta}{\cot \theta}$,we observe that as $\theta \to \frac{\pi }{2}$,the expression takes the indeterminate form $\frac{0}{0}$.
Applying $L'\text{Hospital's rule}$,we differentiate the numerator and the denominator with respect to $\theta$:
$\frac{d}{d\theta} (\frac{\pi}{2} - \theta) = -1$
$\frac{d}{d\theta} (\cot \theta) = -\csc^2 \theta$
Thus,the limit becomes $\mathop {\lim }\limits_{\theta \to \frac{\pi }{2}} \frac{-1}{-\csc^2 \theta} = \mathop {\lim }\limits_{\theta \to \frac{\pi }{2}} \sin^2 \theta$.
Substituting $\theta = \frac{\pi}{2}$,we get $\sin^2(\frac{\pi}{2}) = (1)^2 = 1$.

(C) Given limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}$ which is of the form $\frac{0}{0}$.
Using $L'Hospital's$ rule:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}({e^x} - {e^{\sin x}})}{\frac{d}{dx}(x - \sin x)} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{\sin x}}\cos x}}{{1 - \cos x}}$
Applying $L'Hospital's$ rule again:
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - ({e^{\sin x}}(-\sin x) + \cos x({e^{\sin x}}\cos x))}}{{\sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{\sin x}}\sin x - {e^{\sin x}}\cos^2 x}}{{\sin x}}$
Since $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$ and $\cos x \to 1$,we can simplify the expression using Taylor series expansion for $e^x \approx 1+x+\frac{x^2}{2}$ and $\sin x \approx x - \frac{x^3}{6}$:
$\mathop {\lim }\limits_{x \to 0} \frac{(1+x+\frac{x^2}{2}) - (1+(x-\frac{x^3}{6})+\frac{(x-\frac{x^3}{6})^2}{2})}{x-(x-\frac{x^3}{6})} = \mathop {\lim }\limits_{x \to 0} \frac{1+x+\frac{x^2}{2} - 1 - x + \frac{x^3}{6} - \frac{x^2}{2}}{\frac{x^3}{6}} = \mathop {\lim }\limits_{x \to 0} \frac{\frac{x^3}{6}}{\frac{x^3}{6}} = 1$.

(D) We are given the limit $\mathop {\lim }\limits_{x \to 0} \frac{2}{x}\log (1 + x)$.
Using the property of logarithms,$a \log b = \log b^a$,we can rewrite the expression as $\mathop {\lim }\limits_{x \to 0} 2 \log (1 + x)^{\frac{1}{x}}$.
Since the logarithm function is continuous,we have $2 \log \left( \mathop {\lim }\limits_{x \to 0} (1 + x)^{\frac{1}{x}} \right)$.
We know that $\mathop {\lim }\limits_{x \to 0} (1 + x)^{\frac{1}{x}} = e$.
Therefore,the expression becomes $2 \log_e e = 2 \times 1 = 2$.
Alternatively,using $L'Hospital's$ rule: $\mathop {\lim }\limits_{x \to 0} \frac{2 \log(1+x)}{x} = \mathop {\lim }\limits_{x \to 0} \frac{2 \cdot \frac{1}{1+x}}{1} = \frac{2}{1} = 2$.

(D) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{27^x - 9^x - 3^x + 1}}{{\sqrt 5 - \sqrt {4 + \cos x} }}$.
This is a $\frac{0}{0}$ form.
Applying $L$-Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx}(27^x - 9^x - 3^x + 1) = 27^x \ln 27 - 9^x \ln 9 - 3^x \ln 3 = 3^x \ln 3 (9^x \cdot 3 - 3^x \cdot 2 - 1)$.
Denominator derivative: $\frac{d}{dx}(\sqrt 5 - \sqrt {4 + \cos x}) = -\frac{1}{2\sqrt{4 + \cos x}} \cdot (-\sin x) = \frac{\sin x}{2\sqrt{4 + \cos x}}$.
Thus,$L = \mathop {\lim }\limits_{x \to 0} \frac{27^x \ln 27 - 9^x \ln 9 - 3^x \ln 3}{\frac{\sin x}{2\sqrt{4 + \cos x}}} = \mathop {\lim }\limits_{x \to 0} \frac{2(27^x \ln 27 - 9^x \ln 9 - 3^x \ln 3) \sqrt{4 + \cos x}}{\sin x}$.
Using $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,we multiply and divide by $x$:
$L = 2 \sqrt{4+1} \cdot \mathop {\lim }\limits_{x \to 0} \frac{27^x \ln 27 - 9^x \ln 9 - 3^x \ln 3}{x} = 2\sqrt{5} \cdot (\ln 27 \cdot \ln 27 - \ln 9 \cdot \ln 9 - \ln 3 \cdot \ln 3) = 2\sqrt{5} \cdot (9(\ln 3)^2 - 4(\ln 3)^2 - (\ln 3)^2) = 2\sqrt{5} \cdot 4(\ln 3)^2 = 8\sqrt{5}(\ln 3)^2$.

(B) Using $L$'$H$ôpital's Rule repeatedly,we differentiate the numerator and denominator $n$ times:
$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n \cdot x^{n-1}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n(n-1) \cdot x^{n-2}}}{{{e^x}}} = \dots = \mathop {\lim }\limits_{x \to \infty } \frac{{n!}}{{{e^x}}}$.
Since $\mathop {\lim }\limits_{x \to \infty } e^x = \infty$,we have $\frac{n!}{\infty} = 0$.
This holds true for any whole number $n \ge 0$.

(A) Applying $L$-Hospital's rule to the expression $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}}$,we differentiate the numerator and denominator with respect to $x$:
Numerator: $\frac{d}{dx}(a^x - x^a) = a^x \ln a - a x^{a-1}$
Denominator: $\frac{d}{dx}(x^x - a^a) = x^x(1 + \ln x) - 0 = x^x(1 + \ln x)$
Evaluating the limit as $x \to a$:
$-1 = \frac{a^a \ln a - a(a^{a-1})}{a^a(1 + \ln a)} = \frac{a^a(\ln a - 1)}{a^a(1 + \ln a)} = \frac{\ln a - 1}{\ln a + 1}$
$-1(\ln a + 1) = \ln a - 1$
$-\ln a - 1 = \ln a - 1$
$2 \ln a = 0$ $\Rightarrow \ln a = 0$ $\Rightarrow a = 1$.

(C) Let $y = \mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}}$.
Taking $\log$ on both sides,
$\log y = \mathop {\lim }\limits_{x \to 0} \cot x \log(\cos x)$.
This is of the form $\infty \times 0$. We can rewrite it as:
$\log y = \mathop {\lim }\limits_{x \to 0} \frac{\log(\cos x)}{\tan x}$,which is of the form $\frac{0}{0}$.
Applying $L'\text{Hospital's rule}$,we differentiate the numerator and denominator:
$\log y = \mathop {\lim }\limits_{x \to 0} \frac{\frac{1}{\cos x} \cdot (-\sin x)}{\sec^2 x}$.
$\log y = \mathop {\lim }\limits_{x \to 0} \frac{-\tan x}{\sec^2 x} = \frac{0}{1} = 0$.
Since $\log y = 0$,we have $y = e^0 = 1$.

(C) Given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$.
This is a $\frac{0}{0}$ indeterminate form.
Applying $L'\text{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:
$= \mathop {\lim }\limits_{x \to 0} \frac{{2x \cdot f'({x^2}) - f'(x)}}{{f'(x)}}$.
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{2x \cdot f'({x^2})}{f'(x)} - \frac{f'(x)}{f'(x)} \right)$.
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{2x \cdot f'({x^2})}{f'(x)} - 1 \right)$.
Since $f$ is strictly increasing,$f'(x) > 0$. Assuming $f'(0) \neq 0$,the term $\mathop {\lim }\limits_{x \to 0} \frac{2x \cdot f'({x^2})}{f'(x)} = \frac{0 \cdot f'(0)}{f'(0)} = 0$.
Therefore,the limit is $0 - 1 = -1$.

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to a} \frac{af(x) - xf(a)}{x - a}$,we add and subtract $af(a)$ in the numerator:
$\mathop {\lim }\limits_{x \to a} \frac{af(x) - xf(a) + af(a) - af(a)}{x - a}$
Rearranging the terms:
$\mathop {\lim }\limits_{x \to a} \frac{a[f(x) - f(a)] - f(a)[x - a]}{x - a}$
Splitting the limit:
$a \cdot \mathop {\lim }\limits_{x \to a} \frac{f(x) - f(a)}{x - a} - \mathop {\lim }\limits_{x \to a} f(a)$
Since $f(x)$ is differentiable at $x = a$,$\mathop {\lim }\limits_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$:
$a f'(a) - f(a)$.

(A) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$.
Since the limit is of the form $\frac{0}{0}$ as $x \to 0$,we apply $L'\text{Hospital's rule}$.
Applying the rule once:
$L = \mathop {\lim }\limits_{x \to 0} \frac{2f'(x) - 3 \times 2f'(2x) + 4f'(4x)}{2x} = \mathop {\lim }\limits_{x \to 0} \frac{2f'(x) - 6f'(2x) + 4f'(4x)}{2x}$.
Applying $L'\text{Hospital's rule}$ again:
$L = \mathop {\lim }\limits_{x \to 0} \frac{2f''(x) - 6 \times 2f''(2x) + 4 \times 4f''(4x)}{2} = \mathop {\lim }\limits_{x \to 0} \frac{2f''(x) - 12f''(2x) + 16f''(4x)}{2}$.
As $x \to 0$,$f''(x) \to f''(0) = a$.
$L = \frac{2a - 12a + 16a}{2} = \frac{6a}{2} = 3a$.

(B) Given the limit $L = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$.
Since the form is $\frac{0}{0}$,we apply $L$'$H$ôpital's rule:
$L = \mathop {\lim }\limits_{x \to 1} \frac{\frac{d}{dx}(\sqrt{f(x)} - 1)}{\frac{d}{dx}(\sqrt{x} - 1)}$
$L = \mathop {\lim }\limits_{x \to 1} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}}$
$L = \mathop {\lim }\limits_{x \to 1} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$
Substituting $x = 1$:
$L = \frac{f'(1) \cdot \sqrt{1}}{\sqrt{f(1)}} = \frac{4 \cdot 1}{1} = 4$.

(A) Given the limit $\mathop {\lim }\limits_{x \to a} \frac{f(a)g(x) - f(x)g(a)}{x - a}$.
Substituting $x = a$,we get $\frac{f(a)g(a) - f(a)g(a)}{a - a} = \frac{0}{0}$ form.
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to a} \frac{\frac{d}{dx}(f(a)g(x) - f(x)g(a))}{\frac{d}{dx}(x - a)} = \mathop {\lim }\limits_{x \to a} \frac{f(a)g'(x) - f'(x)g(a)}{1}$.
Now,substitute $x = a$:
$= f(a)g'(a) - f'(a)g(a)$.
Given $f(a) = 2$,$f'(a) = 1$,$g(a) = -3$,$g'(a) = -1$:
$= (2)(-1) - (1)(-3) = -2 + 3 = 1$.

(A) Given the limit: $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$.
Since $f(1) = g(1) = k$,the expression becomes $\lim _{x \rightarrow 1} \frac{k g(x) - k - k f(x) + k}{g(x) - f(x)} = \lim _{x \rightarrow 1} \frac{k(g(x) - f(x))}{g(x) - f(x)}$.
As $x \rightarrow 1$,$g(x) - f(x) \rightarrow g(1) - f(1) = k - k = 0$,which is an indeterminate form of type $\frac{0}{0}$.
Applying $L'\text{Hospital's Rule}$: $\lim _{x \rightarrow 1} \frac{k g'(x) - k f'(x)}{g'(x) - f'(x)} = 4$.
$\lim _{x \rightarrow 1} \frac{k(g'(x) - f'(x))}{g'(x) - f'(x)} = 4$.
Since $g'(1) \neq f'(1)$ (as given that derivatives are not equal),we can cancel the term $(g'(x) - f'(x))$.
Thus,$k = 4$.

(A) We are given the limit $l = \mathop {Lim}\limits_{x \to {0^ + }} x^m (\ln x)^n$.
Rewrite the expression as $l = \mathop {Lim}\limits_{x \to {0^ + }} \frac{(\ln x)^n}{x^{-m}}$.
Since this is an indeterminate form of type $\frac{\infty}{\infty}$ as $x \to 0^+$,we apply $L'\text{Hospital's rule}$ $n$ times.
After applying the rule $n$ times,the numerator becomes $n!$ (multiplied by $(-1)^n$) and the denominator involves $x^m$ multiplied by constants.
Specifically,$l = \mathop {Lim}\limits_{x \to {0^ + }} \frac{n! (-1)^n}{(-m)^n x^{-m}} = \mathop {Lim}\limits_{x \to {0^ + }} \frac{n! (-1)^n x^m}{(-m)^n} = 0$.
Since the limit $l = 0$ regardless of the specific natural numbers $m$ and $n$,the value of $l$ is independent of both $m$ and $n$.

(A) Given the limit $L = \mathop {\text{Limit}}\limits_{x \to 4} \frac{\sqrt{f(x)} - \sqrt{g(x)}}{\sqrt{x} - 2}$.
Since the form is $\frac{\sqrt{2} - \sqrt{2}}{\sqrt{4} - 2} = \frac{0}{0}$,we apply $L'\text{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:
$L = \mathop {\text{Limit}}\limits_{x \to 4} \frac{\frac{d}{dx}(\sqrt{f(x)} - \sqrt{g(x)})}{\frac{d}{dx}(\sqrt{x} - 2)}$
$L = \mathop {\text{Limit}}\limits_{x \to 4} \frac{\frac{f'(x)}{2\sqrt{f(x)}} - \frac{g'(x)}{2\sqrt{g(x)}}}{\frac{1}{2\sqrt{x}}}$
Substitute $x = 4$:
$L = \frac{\frac{f'(4)}{2\sqrt{f(4)}} - \frac{g'(4)}{2\sqrt{g(4)}}}{\frac{1}{2\sqrt{4}}}$
$L = \frac{\frac{9}{2\sqrt{2}} - \frac{6}{2\sqrt{2}}}{\frac{1}{4}}$
$L = \frac{\frac{3}{2\sqrt{2}}}{\frac{1}{4}} = \frac{3}{2\sqrt{2}} \times 4 = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.

(B) Given the limit expression: $L = \mathop {Limit}\limits_{h \to 0} \frac{f(x + 3h) - f(x - 2h)}{h}$.
Since $f(x)$ is differentiable,we can use the definition of the derivative or $L$'Hopital's rule.
Using $L$'Hopital's rule (as it is a $0/0$ form):
$L = \mathop {Limit}\limits_{h \to 0} \frac{\frac{d}{dh}[f(x + 3h) - f(x - 2h)]}{\frac{d}{dh}[h]}$.
Applying the chain rule:
$L = \mathop {Limit}\limits_{h \to 0} \frac{f'(x + 3h) \cdot 3 - f'(x - 2h) \cdot (-2)}{1}$.
$L = \mathop {Limit}\limits_{h \to 0} [3f'(x + 3h) + 2f'(x - 2h)]$.
As $h \to 0$,$f'(x + 3h) \to f'(x)$ and $f'(x - 2h) \to f'(x)$.
Therefore,$L = 3f'(x) + 2f'(x) = 5f'(x)$.

(C) Let $L = \lim_{x \to a} \frac{g(x)f(a) - g(a)f(x)}{x - a}$.
Since the limit is in the form $\frac{0}{0}$ as $x \to a$,we apply $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim_{x \to a} \frac{\frac{d}{dx}[g(x)f(a) - g(a)f(x)]}{\frac{d}{dx}[x - a]}$
$L = \lim_{x \to a} \frac{g'(x)f(a) - g(a)f'(x)}{1}$
Now,substitute $x = a$:
$L = g'(a)f(a) - g(a)f'(a)$
Given $f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2$:
$L = (2)(2) - (-1)(1)$
$L = 4 + 1 = 5$.

(A) Let $L = \lim_{x \to h} \frac{(x + h)f(x) - 2hf(h)}{x - h}$.
Since the limit is in the indeterminate form $\frac{0}{0}$ as $x \to h$,we apply $L'\text{Hospital's rule}$ by differentiating the numerator and the denominator with respect to $x$.
Numerator derivative: $\frac{d}{dx} [(x + h)f(x) - 2hf(h)] = (x + h)f'(x) + f(x)(1) - 0 = (x + h)f'(x) + f(x)$.
Denominator derivative: $\frac{d}{dx} (x - h) = 1$.
Now,evaluate the limit as $x \to h$:
$L = \lim_{x \to h} [(x + h)f'(x) + f(x)] = (h + h)f'(h) + f(h) = 2hf'(h) + f(h)$.
Thus,the correct option is $A$.

(D) Let $f(x) = a \tan^{-1} \frac{\sqrt{x}}{a} - b \tan^{-1} \frac{\sqrt{x}}{b}$ and $g(x) = x\sqrt{x} = x^{3/2}$.
Since the limit is of the form $0/0$,we apply $L$'Hopital's rule:
$\frac{d}{dx} (a \tan^{-1} \frac{\sqrt{x}}{a}) = a \cdot \frac{1}{1 + (\sqrt{x}/a)^2} \cdot \frac{1}{2a\sqrt{x}} = \frac{a^2}{a^2 + x} \cdot \frac{1}{2\sqrt{x}}$.
Similarly,$\frac{d}{dx} (b \tan^{-1} \frac{\sqrt{x}}{b}) = \frac{b^2}{b^2 + x} \cdot \frac{1}{2\sqrt{x}}$.
Derivative of $g(x) = \frac{3}{2}\sqrt{x}$.
Applying the rule: $\lim_{x \to 0^+} \frac{\frac{1}{2\sqrt{x}} (\frac{a^2}{a^2+x} - \frac{b^2}{b^2+x})}{\frac{3}{2}\sqrt{x}} = \lim_{x \to 0^+} \frac{1}{3x} \left( \frac{a^2(b^2+x) - b^2(a^2+x)}{(a^2+x)(b^2+x)} \right)$.
$= \lim_{x \to 0^+} \frac{1}{3x} \left( \frac{a^2b^2 + a^2x - b^2a^2 - b^2x}{(a^2+x)(b^2+x)} \right) = \lim_{x \to 0^+} \frac{1}{3x} \left( \frac{x(a^2 - b^2)}{(a^2+x)(b^2+x)} \right)$.
$= \frac{a^2 - b^2}{3a^2b^2}$.

(C) Given $f(3) = 6$ and $f'(3) = 2$.
We need to evaluate $L = \mathop {\text{Limit}}\limits_{x \to 3} \frac{x f(3) - 3 f(x)}{x - 3}$.
Since the form is $\frac{3 f(3) - 3 f(3)}{3 - 3} = \frac{0}{0}$,we apply $L$'$H$ôpital's rule or algebraic manipulation.
Using $L$'$H$ôpital's rule:
$L = \mathop {\text{Limit}}\limits_{x \to 3} \frac{\frac{d}{dx}(x f(3) - 3 f(x))}{\frac{d}{dx}(x - 3)}$
$L = \mathop {\text{Limit}}\limits_{x \to 3} \frac{f(3) - 3 f'(x)}{1}$
Substituting the values:
$L = f(3) - 3 f'(3) = 6 - 3(2) = 6 - 6 = 0$.
Thus,the correct option is $C$.

(D) $\lim _{x \to \infty} x^{\frac{1}{4}} \sin \frac{1}{\sqrt{x}}$ is of the form $(0 \times \infty)$.
Let $t = \frac{1}{\sqrt{x}}$. As $x \to \infty$,$t \to 0^+$. Then $x^{\frac{1}{4}} = (\frac{1}{t^2})^{\frac{1}{4}} = \frac{1}{\sqrt{t}}$.
$\lim _{t \to 0^+} \frac{\sin t}{\sqrt{t}} = \lim _{t \to 0^+} \sqrt{t} \cdot \frac{\sin t}{t} = 0 \cdot 1 = 0$.
b) $\lim _{x \to \frac{\pi}{2}} (1-\sin x) \tan x$ is of the form $(0 \times \infty)$.
$\lim _{x \to \frac{\pi}{2}} \frac{1-\sin x}{\cot x}$. Applying $L$'Hospital's Rule:
$\lim _{x \to \frac{\pi}{2}} \frac{-\cos x}{-\csc ^{2} x} = \lim _{x \to \frac{\pi}{2}} \cos x \sin ^{2} x = 0 \cdot 1 = 0$.
c) $\lim _{x \to 3^{+}} \frac{[x]^{2}-9}{x^{2}-9}$.
As $x \to 3^+$,$[x] = 3$,so $[x]^2 = 9$.
$\lim _{x \to 3^{+}} \frac{9-9}{x^2-9} = \lim _{x \to 3^{+}} \frac{0}{x^2-9} = 0$.
Since all limits equal $0$,the correct option is $D$.

(D) We are given the displacement function $s(t) = \frac{A}{c^2 - k^2} (\sin kt - \sin ct)$.
To find the limiting value as $c \to k$,we evaluate $\lim_{c \to k} \frac{A(\sin kt - \sin ct)}{c^2 - k^2}$.
Since this is a $0/0$ form,we apply $L$'$H$ôpital's Rule with respect to $c$:
$\lim_{c \to k} \frac{\frac{d}{dc} [A(\sin kt - \sin ct)]}{\frac{d}{dc} [c^2 - k^2]} = \lim_{c \to k} \frac{A(0 - t \cos ct)}{2c}$.
Substituting $c = k$ into the expression,we get:
$\frac{A(-t \cos kt)}{2k} = \frac{-At \cos kt}{2k}$.

(D) Let $L = \mathop {Limit}\limits_{x \to 0} (\cos 2x)^{3/x^2}$.
This is of the form $1^\infty$,so we use the formula $\mathop {Limit}\limits_{x \to a} f(x)^{g(x)} = e^{\mathop {Limit}\limits_{x \to a} g(x)(f(x)-1)}$.
$L = e^{\mathop {Limit}\limits_{x \to 0} \frac{3}{x^2}(\cos 2x - 1)}$.
Using the identity $\cos 2x - 1 = -2\sin^2 x$,we get:
$L = e^{\mathop {Limit}\limits_{x \to 0} \frac{3}{x^2}(-2\sin^2 x)}$.
$L = e^{-6 \mathop {Limit}\limits_{x \to 0} (\frac{\sin x}{x})^2}$.
Since $\mathop {Limit}\limits_{x \to 0} \frac{\sin x}{x} = 1$,we have $L = e^{-6(1)^2} = e^{-6}$.

(C) Let $L = \mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\sin n}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} \right)} \right]^n}$.
As $n \to \infty$,$\frac{\sin n}{n^2} \to 0$ and $\log \left( \frac{en+1}{n+e} \right) \to \log(e) = 1$. Thus,it is a $1^\infty$ form.
Using the formula $\lim_{n \to \infty} [f(n)]^{g(n)} = e^{\lim_{n \to \infty} g(n)(f(n)-1)}$,we have:
$L = e^{\mathop {\lim }\limits_{n \to \infty } n \left[ {\frac{{\sin n}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} \right) - 1} \right]}$
$= e^{\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( {\frac{{en + 1}}{{n + e}}} \right) - n} \right]}$
$= e^{\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( \frac{en+1}{n+e} \right) - n \log e} \right]}$
$= e^{\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( \frac{en+1}{e(n+e)} \right)} \right]}$
$= e^{\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( \frac{en+1}{en+e^2} \right)} \right]}$
$= e^{\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sin n}}{n} + n \log \left( 1 + \frac{1-e^2}{en+e^2} \right)} \right]}$
Since $\frac{\sin n}{n} \to 0$ and $n \log(1+x) \approx nx$ as $x \to 0$:
$= e^{0 + \lim_{n \to \infty} n \cdot \frac{1-e^2}{en+e^2}} = e^{\frac{1-e^2}{e}} = e^{\frac{1}{e} - e}$.

(C) Given limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$
Since the form is $\frac{0}{0}$,we apply $L'\text{Hospital's rule}$:
$= \mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}({e^x} - {e^{ - x}} - 2x)}{\frac{d}{dx}(x - \sin x)} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2}}{{1 - \cos x}}$
Applying $L'\text{Hospital's rule}$ again as it is still $\frac{0}{0}$:
$= \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$
Applying $L'\text{Hospital's rule}$ for the third time:
$= \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{{\cos x}} = \frac{{{e^0} + {e^0}}}{{\cos 0}} = \frac{{1 + 1}}{1} = 2$

(B) The given limit is of the form $\frac{0}{0}$ as $x \to 0$.
Applying $L$'Hopital's rule,we differentiate the numerator and the denominator with respect to $x$.
Using the Leibniz integral rule for the numerator: $\frac{d}{dx} \int_0^x (\tan^{-1} t)^2 dt = (\tan^{-1} x)^2$.
For the denominator: $\frac{d}{dx} (\sin x - x) = \cos x - 1$.
So,the limit becomes $\mathop {\lim }\limits_{x \to 0} \frac{(\tan^{-1} x)^2}{\cos x - 1}$.
Using the standard limits $\mathop {\lim }\limits_{x \to 0} \frac{\tan^{-1} x}{x} = 1$ and $\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$,we rewrite the expression:
$\mathop {\lim }\limits_{x \to 0} \frac{(\tan^{-1} x)^2}{x^2} \cdot \frac{x^2}{\cos x - 1} = (1)^2 \cdot (-2) = -2$.

(C) Let $y = \mathop {\lim }\limits_{x \to \pi /2} \frac{{\int_{\pi /2}^x {t \,dt} }}{{\sin \left( {2x - \pi } \right)}}$.
Using the Fundamental Theorem of Calculus,the integral is $\left[ \frac{t^2}{2} \right]_{\pi/2}^x = \frac{x^2}{2} - \frac{\pi^2}{8}$.
So,$y = \mathop {\lim }\limits_{x \to \pi /2} \frac{\frac{x^2}{2} - \frac{\pi^2}{8}}{\sin(2x - \pi)} = \mathop {\lim }\limits_{x \to \pi /2} \frac{4x^2 - \pi^2}{8 \sin(2x - \pi)}$.
Factoring the numerator: $4x^2 - \pi^2 = (2x - \pi)(2x + \pi)$.
Thus,$y = \frac{1}{8} \mathop {\lim }\limits_{x \to \pi /2} \frac{(2x - \pi)(2x + \pi)}{\sin(2x - \pi)}$.
Using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,where $\theta = 2x - \pi$ as $x \to \pi/2$,we get:
$y = \frac{1}{8} \times (2(\frac{\pi}{2}) + \pi) \times 1 = \frac{1}{8} \times 2\pi = \frac{\pi}{4}$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x {tf(x)dt + \int\limits_0^x xf(t)dt} }}{{{x^2}}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{f(x) \int\limits_0^x t dt + x \int\limits_0^x f(t) dt}}{{{x^2}}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{f(x) \cdot \frac{x^2}{2} + x \int\limits_0^x f(t) dt}}{{{x^2}}}$
$L = \mathop {\lim }\limits_{x \to 0} \left( \frac{f(x)}{2} + \frac{\int\limits_0^x f(t) dt}{x} \right)$
Using $L$'Hopital's rule for the second term: $\mathop {\lim }\limits_{x \to 0} \frac{\int\limits_0^x f(t) dt}{x} = \mathop {\lim }\limits_{x \to 0} \frac{f(x)}{1} = f(0)$
$L = \frac{f(0)}{2} + f(0) = \frac{3}{2} f(0)$
Given $f(0) = 2$,$L = \frac{3}{2} \times 2 = 3$.

(B) Let $t = \{x\}$. As $x \to 1^+$,$t \to 0^+$.
The expression becomes $\mathop {\lim }\limits_{t \to 0^+} \frac{{(1 + t)^{1/t} - e^{1 - t/2}}}{1 - \cos t}$.
Using the expansion $(1 + t)^{1/t} = e^{\frac{1}{t} \ln(1+t)} = e^{\frac{1}{t} (t - t^2/2 + t^3/3 - \dots)} = e^{1 - t/2 + t^2/3 - \dots} = e^{1 - t/2} \cdot e^{t^2/3 - \dots}$.
Substituting this into the limit:
$\mathop {\lim }\limits_{t \to 0^+} \frac{e^{1 - t/2} (e^{t^2/3} - 1)}{t^2/2} = \mathop {\lim }\limits_{t \to 0^+} \frac{e^{1 - t/2} (1 + t^2/3 + \dots - 1)}{t^2/2}$.
$= \mathop {\lim }\limits_{t \to 0^+} \frac{e^{1 - t/2} \cdot t^2/3}{t^2/2} = \frac{e}{3} \cdot 2 = \frac{2e}{3}$.

(B) Let $x = 1 + h$,where $h \to 0$. The expression becomes:
$\mathop {\lim }\limits_{h \to 0} \frac{{\left( {\log \left( {2 + h} \right) - \log 2} \right)\left( {3 \cdot 4^h - 3(1 + h)} \right)}}{{\left( {{{\left( {8 + h} \right)}^{1/3}} - {{\left( {4 + 3h} \right)}^{1/2}}} \right)\sin \pi (1 + h)}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{\log \left( {1 + \frac{h}{2}} \right) \cdot 3(4^h - 1 - h)}}{{2\left[ {{{\left( {1 + \frac{h}{8}} \right)}^{1/3}} - 2{{\left( {1 + \frac{3h}{4}} \right)}^{1/2}}} \right](-\sin \pi h)}}$
Using binomial expansion $(1+x)^n \approx 1+nx$:
$= \mathop {\lim }\limits_{h \to 0} \frac{{\frac{h}{2} \cdot 3 \cdot \frac{h^2 (\ln 4)^2}{2}}}{{2[ (1 + \frac{h}{24}) - 2(1 + \frac{3h}{8}) ](-\pi h)}}$
Simplifying the denominator: $2[1 + \frac{h}{24} - 2 - \frac{3h}{4}] = 2[-1 - \frac{17h}{24}] \approx -2$.
Applying $L$'Hopital's rule or standard limits:
$= \frac{9}{4\pi} (\log 4 - 1)$.

(A) Let $L = \mathop {\lim }\limits_{\theta \to {0^ + }} {(\sin \theta )^{\sin \theta (1 - \sin \theta )}}$.
Taking natural logarithm on both sides,we get $\ln L = \mathop {\lim }\limits_{\theta \to {0^ + }} \sin \theta (1 - \sin \theta ) \ln(\sin \theta)$.
As $\theta \to 0^+$,$\sin \theta \to 0$ and $\ln(\sin \theta) \to -\infty$. This is a $0 \times \infty$ indeterminate form.
We can rewrite this as $\ln L = \mathop {\lim }\limits_{\theta \to {0^ + }} (1 - \sin \theta ) \cdot \frac{\ln(\sin \theta )}{\csc \theta }$.
Applying $L$'$H$ôpital's Rule to the term $\frac{\ln(\sin \theta )}{\csc \theta }$:
$\mathop {\lim }\limits_{\theta \to {0^ + }} \frac{\frac{1}{\sin \theta } \cdot \cos \theta }{-\csc \theta \cot \theta } = \mathop {\lim }\limits_{\theta \to {0^ + }} \frac{\cot \theta }{-\csc \theta \cot \theta } = \mathop {\lim }\limits_{\theta \to {0^ + }} \frac{1}{-\csc \theta } = \mathop {\lim }\limits_{\theta \to {0^ + }} (-\sin \theta ) = 0$.
Thus,$\ln L = (1 - 0) \cdot 0 = 0$.
Therefore,$L = e^0 = 1$.

(A) We evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{e^x - x - 1}{x^2}$.
Substituting $x = 0$,we get $\frac{e^0 - 0 - 1}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}$,which is an indeterminate form.
Applying $L$'$H$ôpital's Rule by differentiating the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(e^x - x - 1)}{\frac{d}{dx}(x^2)} = \mathop {\lim }\limits_{x \to 0} \frac{e^x - 1}{2x}$.
Substituting $x = 0$ again,we get $\frac{e^0 - 1}{2(0)} = \frac{0}{0}$,which is still an indeterminate form.
Applying $L$'$H$ôpital's Rule once more:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(2x)} = \mathop {\lim }\limits_{x \to 0} \frac{e^x}{2}$.
Substituting $x = 0$,we get $\frac{e^0}{2} = \frac{1}{2} = 0.5$.

(D) Let $L = \mathop {\lim }\limits_{x \to 1} {\left( {\frac{4}{\pi }{{\tan }^{ - 1}}x} \right)^{\frac{1}{{({x^2} - 1)}}}}$
Since the form is $1^{\infty}$,we use the formula $\mathop {\lim }\limits_{x \to a} f(x)^{g(x)} = e^{\mathop {\lim }\limits_{x \to a} g(x)(f(x) - 1)}$.
$L = e^{\mathop {\lim }\limits_{x \to 1} \frac{1}{x^2 - 1} \left( \frac{4}{\pi} \tan^{-1} x - 1 \right)}$
$L = e^{\mathop {\lim }\limits_{x \to 1} \frac{4}{\pi} \frac{\tan^{-1} x - \tan^{-1} 1}{x^2 - 1}}$
Using $L$'Hopital's rule or the derivative of $\tan^{-1} x$ at $x=1$:
$L = e^{\frac{4}{\pi} \mathop {\lim }\limits_{x \to 1} \frac{\frac{1}{1+x^2}}{2x}} = e^{\frac{4}{\pi} \cdot \frac{1/2}{2}} = e^{\frac{4}{\pi} \cdot \frac{1}{4}} = e^{\frac{1}{\pi}}$

(C) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{{{(27 + x)}^{\frac{1}{3}}} - 3}}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$
Substituting $x = 0$,we get the indeterminate form $\frac{0}{0}$.
Applying $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}({{(27 + x)}^{\frac{1}{3}}} - 3)}}{{\frac{d}{{dx}}(9 - {{(27 + x)}^{\frac{2}{3}}})}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{3}{{(27 + x)}^{ - \frac{2}{3}}}}}{{ - \frac{2}{3}{{(27 + x)}^{ - \frac{1}{3}}}}}$
$L = \mathop {\lim }\limits_{x \to 0} -\frac{1}{2} {(27 + x)}^{ - \frac{2}{3} + \frac{1}{3}}$
$L = -\frac{1}{2} {(27)}^{ - \frac{1}{3}} = -\frac{1}{2} \times \frac{1}{3} = -\frac{1}{6}$

(B) Given $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$.
First,find the derivative $f'(x)$:
$f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$.
Evaluate $f'(1)$:
$f'(1) = 30(1)^9 - 56(1)^7 + 30(1)^5 - 63(1)^2 + 6(1) = 30 - 56 + 30 - 63 + 6 = -53$.
Now,consider the limit $L = \mathop {\lim }\limits_{\alpha \to 0} \frac{f(1 - \alpha) - f(1)}{\alpha^3 + 3\alpha}$.
Since this is a $\frac{0}{0}$ form,apply $L'\text{Hospital's rule}$:
$L = \mathop {\lim }\limits_{\alpha \to 0} \frac{\frac{d}{d\alpha}(f(1 - \alpha) - f(1))}{\frac{d}{d\alpha}(\alpha^3 + 3\alpha)} = \mathop {\lim }\limits_{\alpha \to 0} \frac{f'(1 - \alpha) \cdot (-1)}{3\alpha^2 + 3}$.
Substitute $\alpha = 0$:
$L = \frac{f'(1) \cdot (-1)}{3(0)^2 + 3} = \frac{-f'(1)}{3} = \frac{-(-53)}{3} = \frac{53}{3}$.

(D) Let $f(x) = \cot^3 x - \tan x$ and $g(x) = \cos(x + \pi/4)$.
At $x = \pi/4$,$f(\pi/4) = 1^3 - 1 = 0$ and $g(\pi/4) = \cos(\pi/2) = 0$. This is a $0/0$ form.
Applying $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx}(\cot^3 x - \tan x) = 3\cot^2 x(-\csc^2 x) - \sec^2 x$.
Denominator derivative: $\frac{d}{dx}(\cos(x + \pi/4)) = -\sin(x + \pi/4)$.
Now,evaluate the limit as $x \to \pi/4$:
$\lim_{x \to \pi/4} \frac{-3\cot^2 x \csc^2 x - \sec^2 x}{-\sin(x + \pi/4)} = \frac{-3(1)^2(\sqrt{2})^2 - (\sqrt{2})^2}{-\sin(\pi/2)} = \frac{-3(2) - 2}{-1} = \frac{-8}{-1} = 8$.

(A) To evaluate the limit $L = \lim_{x \rightarrow 0} \frac{\int_{0}^{x} t \sin(10t) dt}{x}$,we observe that it is of the indeterminate form $\frac{0}{0}$.
Applying $L'H\hat{o}pital's$ rule,we differentiate the numerator and the denominator with respect to $x$.
Using the $Leibniz$ integral rule for the numerator: $\frac{d}{dx} \int_{0}^{x} t \sin(10t) dt = x \sin(10x)$.
The derivative of the denominator $x$ with respect to $x$ is $1$.
Thus,the limit becomes $\lim_{x \rightarrow 0} \frac{x \sin(10x)}{1}$.
As $x \rightarrow 0$,$x \sin(10x) \rightarrow 0 \times \sin(0) = 0 \times 0 = 0$.

(B) We need to evaluate $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{\tan 2x}{x-\frac{\pi}{2}}$.
At $x=\frac{\pi}{2}$,the expression takes the indeterminate form $\frac{0}{0}$.
Let $y = x - \frac{\pi}{2}$. As $x \to \frac{\pi}{2}$,$y \to 0$. Thus,$x = y + \frac{\pi}{2}$.
The limit becomes $\mathop {\lim }\limits_{y \to 0} \frac{\tan 2(y + \frac{\pi}{2})}{y}$.
Since $\tan(2y + \pi) = \tan 2y$,the expression simplifies to $\mathop {\lim }\limits_{y \to 0} \frac{\tan 2y}{y}$.
Using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$,we multiply and divide by $2$:
$\mathop {\lim }\limits_{y \to 0} \frac{\tan 2y}{2y} \times 2 = 1 \times 2 = 2$.

(D) Given $\lim _{x \rightarrow 0} \frac{ae^{x}-b \cos x + ce^{-x}}{x \sin x} = 2.$
Since the denominator $x \sin x \approx x^2$ as $x \rightarrow 0$,the numerator must also approach $0$ at a rate of at least $x^2$ for the limit to exist.
Using Taylor series expansions:
$e^x = 1 + x + \frac{x^2}{2} + \dots$
$\cos x = 1 - \frac{x^2}{2} + \dots$
$e^{-x} = 1 - x + \frac{x^2}{2} + \dots$
Substituting these into the numerator:
$a(1 + x + \frac{x^2}{2}) - b(1 - \frac{x^2}{2}) + c(1 - x + \frac{x^2}{2}) = (a - b + c) + (a - c)x + (\frac{a+b+c}{2})x^2 + \dots$
For the limit to be finite,the coefficients of $x^0$ and $x^1$ must be $0$:
$a - b + c = 0$ and $a - c = 0 \Rightarrow a = c$.
Substituting $a = c$ into the first equation: $a - b + a = 0 \Rightarrow b = 2a$.
The limit becomes $\lim _{x \rightarrow 0} \frac{(\frac{a+b+c}{2})x^2}{x^2} = \frac{a+b+c}{2} = 2$.
Thus,$a + b + c = 4$.

(A) Let $L = \lim_{x \rightarrow a} \frac{(a+2x)^{1/3}-(3x)^{1/3}}{(3a+x)^{1/3}-(4x)^{1/3}}$.
Using $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx} ((a+2x)^{1/3} - (3x)^{1/3}) = \frac{1}{3}(a+2x)^{-2/3}(2) - \frac{1}{3}(3x)^{-2/3}(3) = \frac{2}{3}(a+2x)^{-2/3} - (3x)^{-2/3}$.
Denominator derivative: $\frac{d}{dx} ((3a+x)^{1/3} - (4x)^{1/3}) = \frac{1}{3}(3a+x)^{-2/3} - \frac{1}{3}(4x)^{-2/3}(4) = \frac{1}{3}(3a+x)^{-2/3} - \frac{4}{3}(4x)^{-2/3}$.
Evaluating at $x = a$:
Numerator: $\frac{2}{3}(3a)^{-2/3} - (3a)^{-2/3} = (3a)^{-2/3} (\frac{2}{3} - 1) = -\frac{1}{3}(3a)^{-2/3}$.
Denominator: $\frac{1}{3}(4a)^{-2/3} - \frac{4}{3}(4a)^{-2/3} = (4a)^{-2/3} (\frac{1}{3} - \frac{4}{3}) = -(4a)^{-2/3}$.
Thus,$L = \frac{-\frac{1}{3}(3a)^{-2/3}}{-(4a)^{-2/3}} = \frac{1}{3} \left(\frac{4a}{3a}\right)^{2/3} = \frac{1}{3} \left(\frac{4}{3}\right)^{2/3} = \frac{1}{3} \frac{4^{2/3}}{3^{2/3}} = \frac{4^{2/3}}{3^{5/3}} = \frac{2^{4/3}}{3^{5/3}} = \frac{2 \cdot 2^{1/3}}{3 \cdot 3^{2/3}} = \frac{2}{3} \left(\frac{2}{9}\right)^{1/3}$.

(C) Let $f(x) = x^{2}+bx+c$. Since $\alpha, \beta$ are roots,$f(x) = (x-\alpha)(x-\beta)$.
As $x \rightarrow \beta$,$f(x) \rightarrow 0$.
Using the Taylor series expansion $e^{u} = 1 + u + \frac{u^{2}}{2!} + \dots$,where $u = 2f(x)$:
$\lim _{x \rightarrow \beta} \frac{e^{2f(x)}-1-2f(x)}{(x-\beta)^{2}} = \lim _{x \rightarrow \beta} \frac{(1 + 2f(x) + \frac{(2f(x))^{2}}{2} + \dots) - 1 - 2f(x)}{(x-\beta)^{2}}$
$= \lim _{x \rightarrow \beta} \frac{2(f(x))^{2}}{(x-\beta)^{2}}$
$= \lim _{x \rightarrow \beta} \frac{2((x-\alpha)(x-\beta))^{2}}{(x-\beta)^{2}}$
$= \lim _{x \rightarrow \beta} 2(x-\alpha)^{2} = 2(\beta-\alpha)^{2}$
Since $(\beta-\alpha)^{2} = (\beta+\alpha)^{2} - 4\alpha\beta = (-b)^{2} - 4c = b^{2}-4c$,
The limit is $2(b^{2}-4c)$.

(D) First,calculate $\alpha = \lim_{x \rightarrow \pi/4} \frac{\tan^{3} x - \tan x}{\cos(x + \pi/4)}$. This is a $\frac{0}{0}$ form.
Applying $L'H\hat{o}pital's$ rule:
$\alpha = \lim_{x \rightarrow \pi/4} \frac{3 \tan^{2} x \sec^{2} x - \sec^{2} x}{-\sin(x + \pi/4)} = \frac{3(1)^{2}(2) - 2}{-\sin(\pi/2)} = \frac{6 - 2}{-1} = -4$.
Next,calculate $\beta = \lim_{x \rightarrow 0} (\cos x)^{\cot x}$. This is a $1^{\infty}$ form.
$\beta = e^{\lim_{x \rightarrow 0} \cot x (\cos x - 1)} = e^{\lim_{x \rightarrow 0} \frac{\cos x - 1}{\tan x}} = e^{\lim_{x \rightarrow 0} \frac{-\sin x}{\sec^{2} x}} = e^{0} = 1$.
Since $\alpha = -4$ and $\beta = 1$ are roots of $ax^{2} + bx - 4 = 0$,the sum of roots $\alpha + \beta = -b/a \Rightarrow -4 + 1 = -3 = -b/a \Rightarrow b = 3a$.
The product of roots $\alpha \beta = -4/a \Rightarrow (-4)(1) = -4/a \Rightarrow a = 1$.
Substituting $a = 1$ into $b = 3a$,we get $b = 3$.
Thus,the ordered pair $(a, b)$ is $(1, 3)$.