Concept of limits, Evaluation of algebric limits Questions in English

(B) Let $A = \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}$
$= \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{x + 1}}} \right)^{x + 3}}$
$= \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + \frac{1}{{x + 1}}} \right)}^{x + 1}}} \right]^{\frac{{x + 3}}{{x + 1}}}}$
Since $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{x + 1}}} \right)^{x + 1}} = e$ and $\mathop {\lim }\limits_{x \to \infty } \frac{{x + 3}}{{x + 1}} = 1$,
$A = e^1 = e$.

(C) To evaluate the limit,we rationalize the expression:
$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 1} - x) \times \frac{{\sqrt {{x^2} + 1} + x}}{{\sqrt {{x^2} + 1} + x}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{{({x^2} + 1) - {x^2}}}{{\sqrt {{x^2} + 1} + x}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{1}{{\sqrt {{x^2} + 1} + x}}$
As $x \to \infty$,the denominator $\sqrt {{x^2} + 1} + x \to \infty$.
Therefore,$\frac{1}{\infty} = 0$.

(A) According to the algebra of limits,if $\mathop {\lim }\limits_{x \to a} f(x) = L$ and $\mathop {\lim }\limits_{x \to a} g(x) = M$ exist,then $\mathop {\lim }\limits_{x \to a} [f(x) \cdot g(x)] = L \cdot M$ exists.
Therefore,the existence of individual limits for $f(x)$ and $g(x)$ is a sufficient condition for the existence of the limit of their product.

(A) We use the standard limit formula $\mathop {\lim }\limits_{u \to \infty } (1 + \frac{1}{u})^u = e$.
Given the expression $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{a + bx}}} \right)^{c + dx}}$,we rewrite it as:
$\mathop {\lim }\limits_{x \to \infty } {\left[ {\left( {1 + \frac{1}{{a + bx}}} \right)^{a + bx}} \right]^{\frac{c + dx}{a + bx}}}$
As $x \to \infty$,the term $(1 + \frac{1}{a + bx})^{a + bx} \to e$.
Now,we evaluate the limit of the exponent:
$\mathop {\lim }\limits_{x \to \infty } \frac{c + dx}{a + bx} = \mathop {\lim }\limits_{x \to \infty } \frac{c/x + d}{a/x + b} = \frac{0 + d}{0 + b} = \frac{d}{b}$.
Therefore,the limit is $e^{d/b}$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + \tan x}}{{1 + \sin x}}} \right)^{\text{cosec } x}}$.
This is of the form $1^\infty$,so we use the formula $\mathop {\lim }\limits_{x \to a} f(x)^{g(x)} = e^{\mathop {\lim }\limits_{x \to a} g(x)(f(x) - 1)}$.
$L = e^{\mathop {\lim }\limits_{x \to 0} \text{cosec } x \left( \frac{1 + \tan x}{1 + \sin x} - 1 \right)}$
$L = e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{\sin x} \left( \frac{1 + \tan x - 1 - \sin x}{1 + \sin x} \right)}$
$L = e^{\mathop {\lim }\limits_{x \to 0} \frac{\tan x - \sin x}{\sin x(1 + \sin x)}}$
$L = e^{\mathop {\lim }\limits_{x \to 0} \frac{\sin x (\frac{1}{\cos x} - 1)}{\sin x(1 + \sin x)}}$
$L = e^{\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos x}{\cos x(1 + \sin x)}}$
As $x \to 0$,$\cos x \to 1$ and $\sin x \to 0$.
$L = e^{\frac{1 - 1}{1(1 + 0)}} = e^0 = 1$.

(B) Let $L = \mathop {\lim }\limits_{n \to \infty } {({4^n} + {5^n})^{1/n}}$.
We can factor out the term with the largest base,which is $5^n$:
$L = \mathop {\lim }\limits_{n \to \infty } {\left[ {5^n \left( {{{\left( {\frac{4}{5}} \right)}^n} + 1} \right)} \right]^{1/n}}$
$L = \mathop {\lim }\limits_{n \to \infty } 5 \cdot {\left( {1 + {{\left( {\frac{4}{5}} \right)}^n}} \right)^{1/n}}$
As $n \to \infty$,the term ${\left( {\frac{4}{5}} \right)^n} \to 0$.
Therefore,$L = 5 \cdot {(1 + 0)^0} = 5 \cdot 1 = 5$.

(A) Let $x = \frac{1}{t}$. As $x \to \infty$,$t \to 0^+$.
The expression becomes $\mathop {\lim }\limits_{t \to 0^+} \frac{(\frac{1}{t^2}) \sin(t) - \frac{1}{t}}{1 - |\frac{1}{t}|}$.
Since $x \to \infty$,$t > 0$,so $|\frac{1}{t}| = \frac{1}{t}$.
$= \mathop {\lim }\limits_{t \to 0^+} \frac{\frac{\sin t}{t^2} - \frac{1}{t}}{1 - \frac{1}{t}} = \mathop {\lim }\limits_{t \to 0^+} \frac{\frac{\sin t - t}{t^2}}{\frac{t - 1}{t}} = \mathop {\lim }\limits_{t \to 0^+} \frac{\sin t - t}{t(t - 1)}$.
Using the Taylor series expansion $\sin t = t - \frac{t^3}{3!} + \dots$,we get:
$= \mathop {\lim }\limits_{t \to 0^+} \frac{(t - \frac{t^3}{6} + \dots) - t}{t^2 - t} = \mathop {\lim }\limits_{t \to 0^+} \frac{-\frac{t^3}{6}}{t(t - 1)} = \mathop {\lim }\limits_{t \to 0^+} \frac{-t^2}{6(t - 1)} = \frac{0}{-6} = 0$.

(B) We evaluate the limit: $\mathop {\lim }\limits_{x \to \infty } \sqrt {\frac{{x + \sin x}}{{x - \cos x}}}$
Divide the numerator and denominator inside the square root by $x$:
$= \mathop {\lim }\limits_{x \to \infty } \sqrt {\frac{{1 + \frac{{\sin x}}{x}}}{{1 - \frac{{\cos x}}{x}}}}$
Since $\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x}}{x} = 0$ and $\mathop {\lim }\limits_{x \to \infty } \frac{{\cos x}}{x} = 0$,we have:
$= \sqrt {\frac{{1 + 0}}{{1 - 0}}} = \sqrt {1} = 1$

(C) We have $\mathop {\lim }\limits_{n \to \infty } {({x^n} + {y^n})^{1/n}} = \mathop {\lim }\limits_{n \to \infty } y {\left( 1 + {\left( \frac{x}{y} \right)^n} \right)^{1/n}}$.
Since $0 < x < y$,we have $0 < \frac{x}{y} < 1$.
As $n \to \infty$,${\left( \frac{x}{y} \right)^n} \to 0$.
Therefore,$\mathop {\lim }\limits_{n \to \infty } y {\left( 1 + {\left( \frac{x}{y} \right)^n} \right)^{1/n}} = y {(1 + 0)^0} = y(1) = y$.

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty } (\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})$,we rationalize the expression:
$= \mathop {\lim }\limits_{x \to \infty } \frac{(\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})(\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1})}{\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{({a^2}{x^2} + ax + 1) - ({a^2}{x^2} + 1)}{\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{ax}{\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1}}$
Dividing the numerator and denominator by $x$ (assuming $x > 0$ as $x \to \infty$):
$= \mathop {\lim }\limits_{x \to \infty } \frac{a}{\sqrt {{a^2} + \frac{a}{x} + \frac{1}{{{x^2}}}} + \sqrt {{a^2} + \frac{1}{{{x^2}}}}}$
As $x \to \infty$,$\frac{1}{x} \to 0$ and $\frac{1}{{{x^2}}} \to 0$:
$= \frac{a}{\sqrt {{a^2}} + \sqrt {{a^2}}} = \frac{a}{2|a|}$
If $a > 0$,the limit is $\frac{a}{2a} = \frac{1}{2}$. If $a < 0$,the limit is $\frac{a}{-2a} = -\frac{1}{2}$. Assuming $a > 0$ based on the options,the answer is $\frac{1}{2}$.

(B) Let ${\cos ^{ - 1}}x = y$. As $x \to - 1$,$y \to \pi$.
Then,the limit becomes $\mathop {\lim }\limits_{y \to \pi } \frac{{\sqrt \pi - \sqrt y }}{{\sqrt {1 + \cos y} }}$.
Using the identity $1 + \cos y = 2 \cos^2(y/2)$,we have $\sqrt{1 + \cos y} = \sqrt{2} |\cos(y/2)|$. Since $y \to \pi$,$\cos(y/2) > 0$,so $\sqrt{1 + \cos y} = \sqrt{2} \cos(y/2)$.
Now,$\cos(y/2) = \sin(\pi/2 - y/2)$.
So the limit is $\mathop {\lim }\limits_{y \to \pi } \frac{{\sqrt \pi - \sqrt y }}{{\sqrt{2} \sin(\frac{\pi - y}{2})}}$.
Multiply numerator and denominator by $(\sqrt{\pi} + \sqrt{y})$ and use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$= \mathop {\lim }\limits_{y \to \pi } \frac{\pi - y}{\sqrt{2} \sin(\frac{\pi - y}{2}) (\sqrt{\pi} + \sqrt{y})} = \mathop {\lim }\limits_{y \to \pi } \frac{2 \cdot \frac{\pi - y}{2}}{\sqrt{2} \sin(\frac{\pi - y}{2}) (\sqrt{\pi} + \sqrt{y})} = \frac{2}{\sqrt{2} (2\sqrt{\pi})} = \frac{1}{\sqrt{2\pi}}$.

(B) To evaluate the limit $L = \mathop {\lim }\limits_{x \to \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } \right]$,we rationalize the expression:
$L = \mathop {\lim }\limits_{x \to \infty } \frac{(\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x)(\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x)}{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x}$
$L = \mathop {\lim }\limits_{x \to \infty } \frac{x + \sqrt {x + \sqrt x } - x}{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x} = \mathop {\lim }\limits_{x \to \infty } \frac{\sqrt {x + \sqrt x}}{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x}$
Divide the numerator and denominator by $\sqrt{x}$:
$L = \mathop {\lim }\limits_{x \to \infty } \frac{\sqrt{1 + \frac{1}{\sqrt{x}}}}{\sqrt{1 + \frac{\sqrt{x+\sqrt{x}}}{x}} + 1} = \mathop {\lim }\limits_{x \to \infty } \frac{\sqrt{1 + x^{-1/2}}}{\sqrt{1 + \sqrt{x^{-1} + x^{-3/2}}} + 1}$
As $x \to \infty$,$x^{-1/2} \to 0$,$x^{-1} \to 0$,and $x^{-3/2} \to 0$.
$L = \frac{\sqrt{1+0}}{\sqrt{1+0} + 1} = \frac{1}{1+1} = \frac{1}{2}$.

(C) Given $f(x) = \frac{2}{x - 3}$,$g(x) = \frac{x - 3}{x + 4}$,and $h(x) = - \frac{2(2x + 1)}{x^2 + x - 12}$.
Note that $x^2 + x - 12 = (x - 3)(x + 4)$.
Summing the functions: $f(x) + g(x) + h(x) = \frac{2}{x - 3} + \frac{x - 3}{x + 4} - \frac{2(2x + 1)}{(x - 3)(x + 4)}$.
Taking the common denominator $(x - 3)(x + 4)$:
$= \frac{2(x + 4) + (x - 3)^2 - 2(2x + 1)}{(x - 3)(x + 4)}$
$= \frac{2x + 8 + x^2 - 6x + 9 - 4x - 2}{(x - 3)(x + 4)}$
$= \frac{x^2 - 8x + 15}{(x - 3)(x + 4)}$
$= \frac{(x - 3)(x - 5)}{(x - 3)(x + 4)}$
$= \frac{x - 5}{x + 4}$ for $x \neq 3$.
Therefore,$\lim_{x \to 3} [f(x) + g(x) + h(x)] = \lim_{x \to 3} \frac{x - 5}{x + 4} = \frac{3 - 5}{3 + 4} = - \frac{2}{7}$.

(A) Let $L = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {1 + \sqrt {2 + x} } - \sqrt 3 }}{{x - 2}}$.
Rationalizing the numerator:
$L = \mathop {\lim }\limits_{x \to 2} \frac{(\sqrt {1 + \sqrt {2 + x} } - \sqrt 3)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)}$
$L = \mathop {\lim }\limits_{x \to 2} \frac{1 + \sqrt {2 + x} - 3}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)}$
$L = \mathop {\lim }\limits_{x \to 2} \frac{\sqrt {2 + x} - 2}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)}$
Rationalizing the numerator again:
$L = \mathop {\lim }\limits_{x \to 2} \frac{(\sqrt {2 + x} - 2)(\sqrt {2 + x} + 2)}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)(\sqrt {2 + x} + 2)}$
$L = \mathop {\lim }\limits_{x \to 2} \frac{2 + x - 4}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)(\sqrt {2 + x} + 2)}$
$L = \mathop {\lim }\limits_{x \to 2} \frac{x - 2}{(x - 2)(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3)(\sqrt {2 + x} + 2)}$
$L = \frac{1}{(\sqrt {1 + \sqrt {4} } + \sqrt 3)(\sqrt {4} + 2)} = \frac{1}{(\sqrt 3 + \sqrt 3)(4)} = \frac{1}{2\sqrt 3 \times 4} = \frac{1}{8\sqrt 3}$.

(C) We have $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2 \sin^2(x^2/2)} }}{{2 \sin^2(x/2)}}$.
Since $\sqrt{\sin^2 \theta} = |\sin \theta|$,and for $x \to 0$,$\sin(x^2/2) > 0$,we have:
$= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt{2} \sin(x^2/2)}}{{2 \sin^2(x/2)}} = \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x \to 0} \frac{\sin(x^2/2)}{\sin^2(x/2)}$.
Multiply and divide by $(x^2/2)$ and $(x/2)^2$:
$= \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin(x^2/2)}{x^2/2} \cdot \frac{x^2/2}{(\sin(x/2)/(x/2))^2 \cdot (x^2/4)} \right)$.
$= \frac{1}{\sqrt{2}} \cdot \frac{1/2}{1/4} \cdot \mathop {\lim }\limits_{x \to 0} \frac{1}{1^2} = \frac{1}{\sqrt{2}} \cdot 2 = \sqrt{2}$.

(A) We know that $(1 + x)^{1/x} = e^{\frac{1}{x} \ln(1 + x)}$.
Using the expansion $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,we get:
$(1 + x)^{1/x} = e^{\frac{1}{x}(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots)} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - \dots} = e \cdot e^{-\frac{x}{2} + \frac{x^2}{3} - \dots}$.
Using the expansion $e^u = 1 + u + \frac{u^2}{2!} + \dots$ where $u = -\frac{x}{2} + \frac{x^2}{3} - \dots$:
$(1 + x)^{1/x} = e \left[ 1 + (-\frac{x}{2} + \frac{x^2}{3}) + \frac{1}{2}(-\frac{x}{2})^2 + O(x^3) \right]$
$= e \left[ 1 - \frac{x}{2} + \frac{x^2}{3} + \frac{x^2}{8} + O(x^3) \right] = e \left[ 1 - \frac{x}{2} + \frac{11x^2}{24} + O(x^3) \right]$.
Substituting this into the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{e(1 - \frac{x}{2} + \frac{11x^2}{24}) - e + \frac{ex}{2}}{x^2} = \mathop {\lim }\limits_{x \to 0} \frac{e - \frac{ex}{2} + \frac{11ex^2}{24} - e + \frac{ex}{2}}{x^2} = \mathop {\lim }\limits_{x \to 0} \frac{11ex^2}{24x^2} = \frac{11e}{24}$.

(A) We use the standard limit formula: $\mathop {\lim }\limits_{x \to 0} (1 + f(x))^{1/g(x)} = e^{\mathop {\lim }\limits_{x \to 0} \frac{f(x)}{g(x)}}$.
Given the expression $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}}$,we can rewrite the base as $1 + \left( \frac{1 + 5x^2}{1 + 3x^2} - 1 \right) = 1 + \frac{2x^2}{1 + 3x^2}$.
Thus,the limit becomes $e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x^2} \cdot \frac{2x^2}{1 + 3x^2}}$.
Simplifying the exponent: $\mathop {\lim }\limits_{x \to 0} \frac{2}{1 + 3x^2} = \frac{2}{1 + 0} = 2$.
Therefore,the result is $e^2$.

(D) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty } \frac{(2x - 3)(3x - 4)}{(4x - 5)(5x - 6)}$,we divide the numerator and the denominator by $x^2$:
$= \mathop {\lim }\limits_{x \to \infty } \frac{x(2 - \frac{3}{x}) \cdot x(3 - \frac{4}{x})}{x(4 - \frac{5}{x}) \cdot x(5 - \frac{6}{x})}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{(2 - \frac{3}{x})(3 - \frac{4}{x})}{(4 - \frac{5}{x})(5 - \frac{6}{x})}$
As $x \to \infty$,the terms $\frac{3}{x}, \frac{4}{x}, \frac{5}{x}, \frac{6}{x}$ approach $0$:
$= \frac{(2 - 0)(3 - 0)}{(4 - 0)(5 - 0)} = \frac{2 \times 3}{4 \times 5} = \frac{6}{20} = \frac{3}{10}$

(C) Given $\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{\sin(e^{x-2} - 1)}{\log(x-1)}$.
Let $t = x - 2$. As $x \to 2$,$t \to 0$. Then $x = t + 2$.
Substituting this into the limit:
$\lim_{t \to 0} \frac{\sin(e^t - 1)}{\log(t + 2 - 1)} = \lim_{t \to 0} \frac{\sin(e^t - 1)}{\log(1 + t)}$.
We know that $\lim_{u \to 0} \frac{\sin u}{u} = 1$ and $\lim_{t \to 0} \frac{\log(1 + t)}{t} = 1$.
Multiplying and dividing by $(e^t - 1)$ and $t$:
$\lim_{t \to 0} \left( \frac{\sin(e^t - 1)}{e^t - 1} \right) \times \left( \frac{e^t - 1}{t} \right) \times \left( \frac{t}{\log(1 + t)} \right)$.
Since $\lim_{t \to 0} (e^t - 1) = 0$,the first term approaches $1$.
$\lim_{t \to 0} \frac{e^t - 1}{t} = 1$ and $\lim_{t \to 0} \frac{t}{\log(1 + t)} = 1$.
Therefore,the limit is $1 \times 1 \times 1 = 1$.

(C) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} )$,we rationalize the expression:
$\mathop {\lim }\limits_{x \to \infty } \frac{(\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3})(\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3})}{\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{({x^2} + 8x + 3) - ({x^2} + 4x + 3)}{\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{4x}{\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3}}$
Divide the numerator and denominator by $x$ (for $x > 0$):
$= \mathop {\lim }\limits_{x \to \infty } \frac{4}{\sqrt {1 + \frac{8}{x} + \frac{3}{x^2}} + \sqrt {1 + \frac{4}{x} + \frac{3}{x^2}}}$
As $x \to \infty$,$\frac{1}{x} \to 0$ and $\frac{1}{x^2} \to 0$:
$= \frac{4}{\sqrt{1+0+0} + \sqrt{1+0+0}} = \frac{4}{1+1} = \frac{4}{2} = 2$.

(B) We know the standard limit formula: $\mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$.
Applying this to the given limit: $\mathop {\lim }\limits_{x \to 5} \frac{{{x^k} - {5^k}}}{{x - 5}} = k{(5)^{k - 1}}$.
Given that the limit equals $500$,we have: $k{(5)^{k - 1}} = 500$.
We can rewrite $500$ as $4 \times 125 = 4 \times 5^3 = 4 \times 5^{4-1}$.
Comparing $k{(5)^{k - 1}}$ with $4{(5)^{4 - 1}}$,we get $k = 4$.

(B) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}}$,we rationalize the numerator:
$= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}} \times \frac{{\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} }}{{\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} }}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{(1 - {x^2}) - (1 + {x^2})}}{{{x^2}(\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} )}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{x^2}}}{{{x^2}(\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} )}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{ - 2}}{{\sqrt {1 - {x^2}} + \sqrt {1 + {x^2}} }}$
$= \frac{{ - 2}}{{\sqrt {1 - 0} + \sqrt {1 + 0} }} = \frac{{ - 2}}{{1 + 1}} = -1$

(A) Given the limit: $\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$
Since $x \to -\infty$,we can write $x = -|x| = -\sqrt{x^2}$.
Dividing the numerator and denominator by $|x|$ (where $|x| = -x$ for $x < 0$):
$= \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{{\sqrt {4{x^2} + 5x + 8} }}{{|x|}}}}{{\frac{{4x + 5}}{{|x|}}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4 + \frac{5}{x} + \frac{8}{{{x^2}}}} }}{{ - \left( {4 + \frac{5}{x}} \right)}}$
As $x \to -\infty$,$\frac{1}{x} \to 0$ and $\frac{1}{{{x^2}}} \to 0$.
$= \frac{{\sqrt {4 + 0 + 0} }}{{ - (4 + 0)}} = \frac{2}{{ - 4}} = - \frac{1}{2}$.

(A) Let $y = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{mx}}} \right)^x}$.
We can rewrite the expression as $y = \mathop {\lim }\limits_{x \to \infty } {\left[ {\left( {1 + \frac{1}{{mx}}} \right)^{mx}} \right]^{1/m}}$.
Using the standard limit formula $\mathop {\lim }\limits_{u \to \infty } {\left( {1 + \frac{1}{u}} \right)^u} = e$,where $u = mx$,as $x \to \infty$,$u \to \infty$.
Therefore,$y = e^{1/m}$.

(A) Given limit: $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$.
Substituting $x = 2$,we get the indeterminate form $\frac{0}{0}$.
Using factorization:
$\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}$
$= \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2}$
$= \frac{2^2 + 2(2) + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3$.
Alternatively,applying $L'\text{Hospital's rule}$:
$\lim_{x \to 2} \frac{d/dx(x^3 - 8)}{d/dx(x^2 - 4)} = \lim_{x \to 2} \frac{3x^2}{2x} = \lim_{x \to 2} \frac{3x}{2} = \frac{3(2)}{2} = 3$.

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty} \frac{2x^2 + 3x + 4}{3x^2 + 3x + 4}$,divide the numerator and the denominator by the highest power of $x$,which is $x^2$.
$\mathop {\lim }\limits_{x \to \infty} \frac{2 + \frac{3}{x} + \frac{4}{x^2}}{3 + \frac{3}{x} + \frac{4}{x^2}}$
As $x \to \infty$,the terms $\frac{3}{x}$,$\frac{4}{x^2}$,$\frac{3}{x}$,and $\frac{4}{x^2}$ all approach $0$.
Therefore,the limit is $\frac{2 + 0 + 0}{3 + 0 + 0} = \frac{2}{3}$.

(D) Let $f(x) = \sin \left( \frac{1}{x} \right)$.
For the limit to exist at $x = 0$,the left-hand limit and right-hand limit must be equal.
Left-hand limit: $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} \sin \left( \frac{1}{-h} \right) = \mathop {\lim }\limits_{h \to 0} -\sin \left( \frac{1}{h} \right)$.
As $h \to 0$,$\frac{1}{h} \to \infty$,and $\sin \left( \frac{1}{h} \right)$ oscillates between $-1$ and $1$. Thus,the limit does not exist.
Right-hand limit: $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} \sin \left( \frac{1}{h} \right)$.
Similarly,as $h \to 0$,$\sin \left( \frac{1}{h} \right)$ oscillates between $-1$ and $1$.
Since the left-hand limit and right-hand limit do not approach a unique finite value,$\mathop {\lim }\limits_{x \to 0} \sin \left( \frac{1}{x} \right)$ does not exist.

(B) Let $y = \mathop {\lim }\limits_{x \to 4} \left[ {\frac{{{x^{3/2}} - 8}}{{x - 4}}} \right]$.
Using the formula $\mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$:
Here $n = 3/2$ and $a = 4$.
$y = \frac{3}{2} \times {(4)^{(3/2) - 1}} = \frac{3}{2} \times {(4)^{1/2}} = \frac{3}{2} \times 2 = 3$.
Alternatively,applying $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 4} \frac{{\frac{d}{{dx}}({x^{3/2}} - 8)}}{{\frac{d}{{dx}}(x - 4)}} = \mathop {\lim }\limits_{x \to 4} \frac{{\frac{3}{2}{x^{1/2}}}}{1} = \frac{3}{2} \times \sqrt{4} = \frac{3}{2} \times 2 = 3$.

(D) Given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}}}}{{{e^{\left( {\frac{1}{x} + 1} \right)}}}}$
Using the property of exponents,$e^{a+b} = e^a \cdot e^b$,we can rewrite the denominator:
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}}}}{{{e^{1/x}} \cdot e^1}}$
Canceling the common term $e^{1/x}$ (which is non-zero for all $x \neq 0$):
$\mathop {\lim }\limits_{x \to 0} \frac{1}{e} = \frac{1}{e} = e^{-1}$
Since $e^{-1}$ is not equal to $0$,$1$,or does not exist,the correct option is $D$.

(D) We have $\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{n}{{n + y}}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{1}{1 + \frac{y}{n}}} \right)^n}$
$= \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{y}{n}} \right)^{-n}}$
$= \mathop {\lim }\limits_{n \to \infty } {\left[ {\left( {1 + \frac{y}{n}} \right)^n} \right]^{-1}}$
Since $\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{y}{n}} \right)^n} = e^y$,we get:
$= (e^y)^{-1} = e^{-y}$.

(A) To find $\mathop {\lim }\limits_{x \to 0} f(x)$,we evaluate the left-hand limit $(L.H.L.)$ and the right-hand limit $(R.H.L.)$.
$L.H.L. = \mathop {\lim }\limits_{x \to 0^-} f(x) = \mathop {\lim }\limits_{x \to 0^-} (x) = 0$.
$R.H.L. = \mathop {\lim }\limits_{x \to 0^+} f(x) = \mathop {\lim }\limits_{x \to 0^+} (x^2) = 0^2 = 0$.
Since $L.H.L. = R.H.L. = 0$,the limit exists and is equal to $0$.

(C) Given the limit: $\mathop {\lim }\limits_{x \to 2} \frac{{{3^{x/2}} - 3}}{{{3^x} - 9}}$
We can rewrite the denominator as a difference of squares: ${3^x} - 9 = {({3^{x/2}})^2} - {3^2} = ({3^{x/2}} - 3)({3^{x/2}} + 3)$
Substituting this into the limit expression:
$\mathop {\lim }\limits_{x \to 2} \frac{{{3^{x/2}} - 3}}{({3^{x/2}} - 3)({3^{x/2}} + 3)}$
Canceling the common term $({3^{x/2}} - 3)$:
$\mathop {\lim }\limits_{x \to 2} \frac{1}{{{3^{x/2}} + 3}}$
Evaluating the limit as $x \to 2$:
$\frac{1}{{{3^{2/2}} + 3}} = \frac{1}{{{3^1} + 3}} = \frac{1}{3 + 3} = \frac{1}{6}$

(C) We use the standard limit formula $\mathop {\lim }\limits_{x \to \infty } {(1 + \frac{a}{x})^x} = e^a$.
Given expression: $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2 - 5}}{{x + 2}}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{5}{{x + 2}}} \right)^x}$.
Rewrite the expression as: $\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 - \frac{5}{{x + 2}}} \right)}^{\frac{{x + 2}}{{-5}}}}} \right]^{\frac{-5x}{x + 2}}}$.
Since $\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{5}{{x + 2}}} \right)^{\frac{x + 2}{-5}}} = e$ and $\mathop {\lim }\limits_{x \to \infty } \frac{-5x}{x + 2} = \mathop {\lim }\limits_{x \to \infty } \frac{-5}{1 + \frac{2}{x}} = -5$.
The limit is $e^{-5}$.

(C) We know the Taylor series expansion for $e^x$ is $e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots$
Substituting this into the limit expression:
$\mathop {\lim }\limits_{x \to 0} \frac{(1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots) - 1}{x}$
$= \mathop {\lim }\limits_{x \to 0} \frac{\frac{x}{1!} + \frac{x^2}{2!} + \dots}{x}$
$= \mathop {\lim }\limits_{x \to 0} (1 + \frac{x}{2!} + \dots) = 1$.

(D) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \left[ \frac{\sqrt{a + x} - \sqrt{a - x}}{x} \right]$,we rationalize the numerator:
$\mathop {\lim }\limits_{x \to 0} \left[ \frac{(\sqrt{a + x} - \sqrt{a - x})(\sqrt{a + x} + \sqrt{a - x})}{x(\sqrt{a + x} + \sqrt{a - x})} \right]$
$= \mathop {\lim }\limits_{x \to 0} \left[ \frac{(a + x) - (a - x)}{x(\sqrt{a + x} + \sqrt{a - x})} \right]$
$= \mathop {\lim }\limits_{x \to 0} \left[ \frac{2x}{x(\sqrt{a + x} + \sqrt{a - x})} \right]$
$= \mathop {\lim }\limits_{x \to 0} \left[ \frac{2}{\sqrt{a + x} + \sqrt{a - x}} \right]$
$= \frac{2}{\sqrt{a + 0} + \sqrt{a - 0}} = \frac{2}{2\sqrt{a}} = \frac{1}{\sqrt{a}}$.

(A) Let $f(x) = (1 + x)^{1/x}$.
Using the expansion $\log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,we have:
$(1 + x)^{1/x} = e^{\frac{1}{x} \log(1 + x)} = e^{\frac{1}{x} (x - \frac{x^2}{2} + \frac{x^3}{3} - \dots)} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - \dots} = e \cdot e^{-\frac{x}{2} + \frac{x^2}{3} - \dots}$.
Using the expansion $e^u = 1 + u + \frac{u^2}{2!} + \dots$,where $u = -\frac{x}{2} + \frac{x^2}{3} - \dots$:
$(1 + x)^{1/x} = e \left( 1 + (-\frac{x}{2} + \frac{x^2}{3} - \dots) + \frac{(-\frac{x}{2} + \dots)^2}{2} + \dots \right) = e - \frac{ex}{2} + O(x^2)$.
Thus,$\mathop {\lim }\limits_{x \to 0} \frac{(1 + x)^{1/x} - e}{x} = \mathop {\lim }\limits_{x \to 0} \frac{(e - \frac{ex}{2} + O(x^2)) - e}{x} = \mathop {\lim }\limits_{x \to 0} (-\frac{e}{2} + O(x)) = -\frac{e}{2}$.

(D) We use the limit formula $\mathop {\lim }\limits_{u \to 0} {(1 + u)^{1/u}} = e$.
Let $L = \mathop {\lim }\limits_{m \to \infty } {\left( {\cos \frac{x}{m}} \right)^m}$.
Using the identity $\cos \theta = 1 - 2\sin^2(\theta/2)$,we have:
$L = \mathop {\lim }\limits_{m \to \infty } {\left[ {1 - 2{{\sin }^2}\left( {\frac{x}{{2m}}} \right)} \right]^m}$
$L = \exp \left( \mathop {\lim }\limits_{m \to \infty } m \cdot \left( -2\sin^2\left( \frac{x}{2m} \right) \right) \right)$
Using $\sin \theta \approx \theta$ as $\theta \to 0$:
$L = \exp \left( \mathop {\lim }\limits_{m \to \infty } m \cdot \left( -2 \cdot \frac{x^2}{4m^2} \right) \right)$
$L = \exp \left( \mathop {\lim }\limits_{m \to \infty } -\frac{x^2}{2m} \right) = e^0 = 1$.

(C) We have the limit $\mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{x + a}}{{x + b}}} \right)^{x + b}}$.
Rewrite the expression inside the limit as $\mathop {\lim }\limits_{x \to \infty } \,{\left( {1 + \frac{{a - b}}{{x + b}}} \right)^{x + b}}$.
Using the standard limit formula $\mathop {\lim }\limits_{u \to \infty } (1 + \frac{k}{u})^u = e^k$,where $u = x + b$ and $k = a - b$:
$\mathop {\lim }\limits_{x \to \infty } \,{\left( {1 + \frac{{a - b}}{{x + b}}} \right)^{x + b}} = e^{a - b}$.
Thus,the correct option is $C$.

(A) Let $L = \mathop {\lim }\limits_{x \to \pi /2} \frac{{{a^{\cot x}} - {a^{\cos x}}}}{{\cot x - \cos x}}$.
We can factor out ${a^{\cos x}}$ from the numerator:
$L = \mathop {\lim }\limits_{x \to \pi /2} \frac{{{a^{\cos x}}({a^{\cot x - \cos x}} - 1)}}{{\cot x - \cos x}}$.
As $x \to \pi /2$,$\cos x \to 0$,so ${a^{\cos x}} \to {a^0} = 1$.
Using the standard limit $\mathop {\lim }\limits_{u \to 0} \frac{{{a^u} - 1}}{u} = \log a$,where $u = \cot x - \cos x$ (note that as $x \to \pi /2$,$u \to 0 - 0 = 0$):
$L = 1 \times \log a = \log a$.

(C) To find the limit $\mathop {\lim }\limits_{x \to 3} [x]$,we evaluate the left-hand limit and the right-hand limit.
Right-hand limit: $\mathop {\lim }\limits_{h \to 0^+} [3 + h] = 3$.
Left-hand limit: $\mathop {\lim }\limits_{h \to 0^+} [3 - h] = 2$.
Since the left-hand limit $(2)$ is not equal to the right-hand limit $(3)$,the limit $\mathop {\lim }\limits_{x \to 3} [x]$ does not exist.

(C) The limit $\mathop {\lim }\limits_{x \to 0} \cos \frac{1}{x}$ involves an oscillating function.
As $x \to 0$,the value of $\frac{1}{x}$ approaches $\infty$.
The function $\cos \frac{1}{x}$ oscillates infinitely many times between $-1$ and $1$ as $x$ approaches $0$.
Since the function does not approach a unique finite value,the limit does not exist.

(B) Let $L = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} \right)^{n(n - 1)}}$.
We can rewrite the expression as $L = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{(n^2 - n - 1) + 2}}{{n^2 - n - 1}}} \right)^{n(n - 1)}} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{2}{{n^2 - n - 1}}} \right)^{n(n - 1)}}$.
Using the standard limit formula $\mathop {\lim }\limits_{x \to \infty } {(1 + \frac{a}{x})^x} = e^a$,we have:
$L = \exp \left( \mathop {\lim }\limits_{n \to \infty } \frac{2}{n^2 - n - 1} \cdot n(n - 1) \right)$.
$L = \exp \left( \mathop {\lim }\limits_{n \to \infty } \frac{2(n^2 - n)}{n^2 - n - 1} \right)$.
Dividing numerator and denominator by $n^2$,we get $\exp(2) = e^2$.

(A) We are given the limit: $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 3}}{{x + 1}}} \right)^{x + 1}}$
Rewrite the expression inside the parenthesis: $\frac{x+3}{x+1} = \frac{x+1+2}{x+1} = 1 + \frac{2}{x+1}$
Now the limit becomes: $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{2}{x+1}} \right)^{x+1}}$
Using the standard limit formula $\mathop {\lim }\limits_{u \to \infty } (1 + \frac{k}{u})^u = e^k$,where $u = x+1$ and $k = 2$:
$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{2}{x+1}} \right)^{x+1}} = e^2$

(B) We know that $\mathop {\lim }\limits_{x \to 0} {(1 + x)^{\frac{1}{x}}} = e$.
Given the expression $\mathop {\lim }\limits_{x \to 0} {(1 - ax)^{\frac{1}{x}}}$,we can rewrite it as $\mathop {\lim }\limits_{x \to 0} {[(1 + (-ax))^{\frac{1}{-ax}}]^{\frac{-ax}{x}}}$.
Since $\mathop {\lim }\limits_{x \to 0} {(1 + (-ax))^{\frac{1}{-ax}}} = e$,the expression simplifies to $e^{-a}$.

(B) We are given the limit $L = \mathop {\lim }\limits_{n \to \infty } {({3^n} + {4^n})^{\frac{1}{n}}}$.
Factor out the term with the largest base,which is $4^n$:
$L = \mathop {\lim }\limits_{n \to \infty } {\left[ {4^n \left( {\left( \frac{3}{4} \right)^n + 1} \right)} \right]^{\frac{1}{n}}}$
$L = \mathop {\lim }\limits_{n \to \infty } 4 \cdot {\left[ {1 + \left( \frac{3}{4} \right)^n} \right]^{\frac{1}{n}}}$
As $n \to \infty$,the term $\left( \frac{3}{4} \right)^n \to 0$.
Therefore,$L = 4 \cdot (1 + 0)^0 = 4 \cdot 1 = 4$.

(B) We know that $\mathop {\lim }\limits_{x \to \infty } {(1 + f(x))^{g(x)}} = e^{\mathop {\lim }\limits_{x \to \infty } f(x)g(x)}$ if $\mathop {\lim }\limits_{x \to \infty } f(x) = 0$.
Given $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)^{2x}} = {e^2}$.
Applying the formula,we get $e^{\mathop {\lim }\limits_{x \to \infty } (\frac{a}{x} + \frac{b}{{{x^2}}}) \cdot 2x} = {e^2}$.
This simplifies to $e^{\mathop {\lim }\limits_{x \to \infty } (2a + \frac{2b}{x})} = {e^2}$.
Evaluating the limit,we get $e^{2a + 0} = {e^2}$.
Thus,$e^{2a} = {e^2}$,which implies $2a = 2$,so $a = 1$.
Since the term $\frac{b}{x^2}$ vanishes as $x \to \infty$ regardless of the value of $b$,$b$ can be any real number.
Therefore,$a = 1$ and $b \in \mathbb{R}$.

(B) We use the standard limit formula: $\mathop {\lim }\limits_{x \to \infty } (1 + \frac{a}{x})^x = e^a$.
Given expression: $\mathop {\lim }\limits_{x \to \infty } (1 - \frac{4}{x - 1})^{3x - 1}$.
Let $t = x - 1$,then as $x \to \infty$,$t \to \infty$.
Substituting $x = t + 1$:
$\mathop {\lim }\limits_{t \to \infty } (1 - \frac{4}{t})^{3(t + 1) - 1} = \mathop {\lim }\limits_{t \to \infty } (1 - \frac{4}{t})^{3t + 2}$.
This can be written as: $\mathop {\lim }\limits_{t \to \infty } [(1 - \frac{4}{t})^t]^3 \times \mathop {\lim }\limits_{t \to \infty } (1 - \frac{4}{t})^2$.
$= (e^{-4})^3 \times (1 - 0)^2 = e^{-12} \times 1 = e^{-12}$.

(D) To find the limit $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 4x + 3}}$,we first factorize the numerator and the denominator.
Numerator: $x^2 + 3x + 2 = (x + 1)(x + 2)$
Denominator: $x^2 + 4x + 3 = (x + 1)(x + 3)$
Now,substitute these into the limit expression:
$\mathop {\lim }\limits_{x \to - 1} \frac{(x + 1)(x + 2)}{(x + 1)(x + 3)}$
Since $x \to - 1$,$x + 1 \neq 0$,we can cancel the common factor $(x + 1)$:
$\mathop {\lim }\limits_{x \to - 1} \frac{x + 2}{x + 3}$
Now,evaluate the limit by substituting $x = - 1$:
$\frac{- 1 + 2}{- 1 + 3} = \frac{1}{2}$.

(B) We use the standard limit formula $\mathop {\lim }\limits_{x \to \infty } (1 + \frac{a}{x})^{bx} = e^{ab}$.
First,rewrite the expression as $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{-6}{3x + 2}} \right)^{\frac{x + 1}{3}}}$.
This is of the form $1^{\infty}$,so the limit is $e^{\mathop {\lim }\limits_{x \to \infty } (\frac{-6}{3x + 2} \times \frac{x + 1}{3})}$.
Calculate the exponent limit: $\mathop {\lim }\limits_{x \to \infty } \frac{-6(x + 1)}{3(3x + 2)} = \mathop {\lim }\limits_{x \to \infty } \frac{-6x - 6}{9x + 6} = -\frac{6}{9} = -\frac{2}{3}$.
Thus,the final value is $e^{-2/3}$.

(D) We are given the limit: $\mathop {\lim }\limits_{x \to \infty } \frac{(x + 1)(3x + 4)}{x^2(x - 8)}$
Expand the numerator: $(x + 1)(3x + 4) = 3x^2 + 4x + 3x + 4 = 3x^2 + 7x + 4$
Expand the denominator: $x^2(x - 8) = x^3 - 8x^2$
Now,divide both the numerator and the denominator by the highest power of $x$ in the denominator,which is $x^3$:
$\mathop {\lim }\limits_{x \to \infty } \frac{\frac{3x^2}{x^3} + \frac{7x}{x^3} + \frac{4}{x^3}}{\frac{x^3}{x^3} - \frac{8x^2}{x^3}} = \mathop {\lim }\limits_{x \to \infty } \frac{\frac{3}{x} + \frac{7}{x^2} + \frac{4}{x^3}}{1 - \frac{8}{x}}$
As $x \to \infty$,the terms $\frac{3}{x}, \frac{7}{x^2}, \frac{4}{x^3}$,and $\frac{8}{x}$ all approach $0$.
Therefore,the limit is $\frac{0 + 0 + 0}{1 - 0} = \frac{0}{1} = 0$.