Concept of limits, Evaluation of algebric limits Questions in English

(A) regular $n$-gon inscribed in a circle of radius $R$ has a perimeter given by $P_n = 2nR \sin(\frac{\pi}{n})$.
As $n \to \infty$,we use the limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
Substituting $x = \frac{\pi}{n}$,as $n \to \infty$,$x \to 0$.
Thus,$\lim_{n \to \infty} P_n = \lim_{n \to \infty} 2nR \sin(\frac{\pi}{n}) = \lim_{n \to \infty} 2\pi R \frac{\sin(\pi/n)}{\pi/n} = 2\pi R(1) = 2\pi R$.
Therefore,the perimeter approaches the circumference of the circle,which is $2\pi R$.

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{x({e^x} - 1)}}{{1 - \cos x}}$.
Using the standard limits $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = 1$ and the identity $1 - \cos x = 2 \sin^2(x/2)$:
$\mathop {\lim }\limits_{x \to 0} \frac{{x({e^x} - 1)}}{{2 \sin^2(x/2)}}$.
Multiply and divide by $x$ and $(x/2)^2$ to use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{{{e^x} - 1}}{x} \right) \cdot \frac{x^2}{2 \sin^2(x/2)}$.
$= 1 \cdot \mathop {\lim }\limits_{x \to 0} \frac{x^2}{2 (x/2)^2 \cdot (\frac{\sin(x/2)}{x/2})^2}$.
$= \mathop {\lim }\limits_{x \to 0} \frac{x^2}{2 (x^2/4) \cdot 1^2} = \frac{1}{2/4} = 2$.

(D) To find the limit $\mathop {\lim }\limits_{x \to 1} \frac{1}{|1 - x|}$,we examine the left-hand and right-hand limits.
Left-hand limit: $\mathop {\lim }\limits_{x \to 1^-} \frac{1}{|1 - x|} = \mathop {\lim }\limits_{h \to 0^+} \frac{1}{|1 - (1 - h)|} = \mathop {\lim }\limits_{h \to 0^+} \frac{1}{|h|} = \infty$.
Right-hand limit: $\mathop {\lim }\limits_{x \to 1^+} \frac{1}{|1 - x|} = \mathop {\lim }\limits_{h \to 0^+} \frac{1}{|1 - (1 + h)|} = \mathop {\lim }\limits_{h \to 0^+} \frac{1}{|-h|} = \mathop {\lim }\limits_{h \to 0^+} \frac{1}{h} = \infty$.
Since both one-sided limits approach $\infty$,the limit is $\infty$.

(C) We need to evaluate the limit: $\mathop {\lim }\limits_{n \to \infty } \frac{{n{{(2n + 1)}^2}}}{{(n + 2)({n^2} + 3n - 1)}}$.
Expanding the numerator: $n(2n+1)^2 = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.
Expanding the denominator: $(n+2)(n^2 + 3n - 1) = n^3 + 3n^2 - n + 2n^2 + 6n - 2 = n^3 + 5n^2 + 5n - 2$.
Now,the limit becomes: $\mathop {\lim }\limits_{n \to \infty } \frac{{4n^3 + 4n^2 + n}}{{n^3 + 5n^2 + 5n - 2}}$.
Divide both the numerator and the denominator by $n^3$:
$= \mathop {\lim }\limits_{n \to \infty } \frac{{4 + \frac{4}{n} + \frac{1}{n^2}}}{{1 + \frac{5}{n} + \frac{5}{n^2} - \frac{2}{n^3}}}$.
As $n \to \infty$,all terms with $n$ in the denominator approach $0$:
$= \frac{{4 + 0 + 0}}{{1 + 0 + 0 - 0}} = 4$.

(B) To evaluate the limit $\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }}$,divide both the numerator and the denominator by $\sqrt{n}$:
$\mathop {\lim }\limits_{n \to \infty } \frac{\frac{\sqrt{n}}{\sqrt{n}}}{\frac{\sqrt{n}}{\sqrt{n}} + \frac{\sqrt{n+1}}{\sqrt{n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{1 + \sqrt{1 + \frac{1}{n}}}$
As $n \to \infty$,the term $\frac{1}{n} \to 0$.
Therefore,the limit becomes $\frac{1}{1 + \sqrt{1 + 0}} = \frac{1}{1 + 1} = \frac{1}{2}$.

(A) To find $\lim_{x \to 1} f(x)$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$ at $x = 1$.
Left-hand limit $(LHL)$:
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x) = 1$
Right-hand limit $(RHL)$:
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2 - x) = 2 - 1 = 1$
Since the left-hand limit equals the right-hand limit,the limit exists and is equal to $1$.

(B) We know the standard limit formula: $\mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n \cdot a^{n-1}$.
Given the expression: $\mathop {\lim }\limits_{x \to 2} \frac{{{x^n} - {2^n}}}{{x - 2}} = 80$.
Applying the formula with $a = 2$,we get: $n \cdot 2^{n-1} = 80$.
We can write $80$ as $5 \cdot 16 = 5 \cdot 2^4$.
Comparing $n \cdot 2^{n-1} = 5 \cdot 2^4$,we find $n = 5$.

(C) We use the standard limit formula: $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x}} \right)^x} = e^a$.
Comparing this with the given expression $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{2}{x}} \right)^x}$,we have $a = 2$.
Therefore,the limit is $e^2$.

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3}$.
First,factor the denominator: $2x^2 + x - 3 = (2x + 3)(x - 1)$.
Substitute this into the limit: $\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{(2x + 3)(x - 1)}$.
Since $(x - 1) = (\sqrt{x} - 1)(\sqrt{x} + 1)$,the expression becomes: $\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{(2x + 3)(\sqrt{x} - 1)(\sqrt{x} + 1)}$.
Cancel the common factor $(\sqrt{x} - 1)$: $\mathop {\lim }\limits_{x \to 1} \frac{2x - 3}{(2x + 3)(\sqrt{x} + 1)}$.
Now,substitute $x = 1$: $\frac{2(1) - 3}{(2(1) + 3)(\sqrt{1} + 1)} = \frac{-1}{(5)(2)} = -\frac{1}{10}$.

(D) Let $f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1}$.
For the right-hand limit $(x \to 0^+)$:
$\mathop {\lim }\limits_{x \to 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \mathop {\lim }\limits_{h \to 0} \frac{e^{1/h} - 1}{e^{1/h} + 1} = \mathop {\lim }\limits_{h \to 0} \frac{1 - e^{-1/h}}{1 + e^{-1/h}} = \frac{1 - 0}{1 + 0} = 1$.
For the left-hand limit $(x \to 0^-)$:
$\mathop {\lim }\limits_{x \to 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \mathop {\lim }\limits_{h \to 0} \frac{e^{-1/h} - 1}{e^{-1/h} + 1} = \frac{0 - 1}{0 + 1} = -1$.
Since the right-hand limit $(1)$ is not equal to the left-hand limit $(-1)$,the limit does not exist.

(D) The limit exists if and only if the left-hand limit and the right-hand limit are equal.
For the left-hand limit: $\mathop {\lim }\limits_{x \to 0^-} \frac{|x|}{x} = \mathop {\lim }\limits_{x \to 0^-} \frac{-x}{x} = -1$.
For the right-hand limit: $\mathop {\lim }\limits_{x \to 0^+} \frac{|x|}{x} = \mathop {\lim }\limits_{x \to 0^+} \frac{x}{x} = 1$.
Since the left-hand limit $(-1)$ is not equal to the right-hand limit $(1)$,the limit does not exist.

(C) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty } \sqrt x (\sqrt {x + 5} - \sqrt x )$,we rationalize the expression:
$\mathop {\lim }\limits_{x \to \infty } \sqrt x (\sqrt {x + 5} - \sqrt x ) \times \frac{\sqrt {x + 5} + \sqrt x}{\sqrt {x + 5} + \sqrt x}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{\sqrt x (x + 5 - x)}{\sqrt {x + 5} + \sqrt x}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{5\sqrt x}{\sqrt x(\sqrt {1 + 5/x} + 1)}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{5}{\sqrt {1 + 5/x} + 1}$
As $x \to \infty$,$5/x \to 0$,so the expression becomes $\frac{5}{\sqrt{1+0} + 1} = \frac{5}{2}$.

(C) Given limit: $\mathop {\lim }\limits_{x \to 1} \frac{x - 1}{2x^2 - 7x + 5}$.
Substituting $x = 1$,we get the indeterminate form $\frac{0}{0}$.
Factorizing the denominator: $2x^2 - 7x + 5 = (x - 1)(2x - 5)$.
Thus,$\mathop {\lim }\limits_{x \to 1} \frac{x - 1}{(x - 1)(2x - 5)} = \mathop {\lim }\limits_{x \to 1} \frac{1}{2x - 5}$.
Substituting $x = 1$: $\frac{1}{2(1) - 5} = \frac{1}{-3} = -\frac{1}{3}$.
Alternatively,using $L$-Hospital's rule: $\mathop {\lim }\limits_{x \to 1} \frac{\frac{d}{dx}(x - 1)}{\frac{d}{dx}(2x^2 - 7x + 5)} = \mathop {\lim }\limits_{x \to 1} \frac{1}{4x - 7} = \frac{1}{4(1) - 7} = -\frac{1}{3}$.

(A) We are evaluating the limit $\mathop {\lim }\limits_{x \to n + 0} (x - [x])$.
Since $x \to n + 0$,$x$ is slightly greater than $n$.
For any $x$ such that $n < x < n + 1$,the greatest integer function $[x] = n$.
Therefore,$\mathop {\lim }\limits_{x \to n + 0} (x - [x]) = \mathop {\lim }\limits_{x \to n + 0} (x - n) = n - n = 0$.

(D) Let $f(x) = \frac{x}{|x| + x^2}$.
For the left-hand limit $(LHL)$:
$\mathop {\lim }\limits_{x \to 0^-} f(x) = \mathop {\lim }\limits_{h \to 0} \frac{-h}{|-h| + (-h)^2} = \mathop {\lim }\limits_{h \to 0} \frac{-h}{h + h^2} = \mathop {\lim }\limits_{h \to 0} \frac{-1}{1 + h} = -1$.
For the right-hand limit $(RHL)$:
$\mathop {\lim }\limits_{x \to 0^+} f(x) = \mathop {\lim }\limits_{h \to 0} \frac{h}{|h| + h^2} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h + h^2} = \mathop {\lim }\limits_{h \to 0} \frac{1}{1 + h} = 1$.
Since $LHL \neq RHL$,the limit does not exist.

(C) We use the algebraic identity $x^2 - a^2 = (x - a)(x + a)$.
$\mathop {\lim }\limits_{x \to a} \frac{{{x^2} - {a^2}}}{{x - a}} = \mathop {\lim }\limits_{x \to a} \frac{(x - a)(x + a)}{x - a}$
Since $x \to a$,$x - a \neq 0$,so we can cancel the term $(x - a)$.
$= \mathop {\lim }\limits_{x \to a} (x + a) = a + a = 2a.$

(A) We use the standard limit formula: $\mathop {\lim }\limits_{x \to a} \frac{{x^n - a^n}}{{x - a}} = na^{n-1}$.
Let $u = x + 2$. As $x \to a$,$u \to a + 2$.
The expression becomes $\mathop {\lim }\limits_{u \to a+2} \frac{{u^{5/3} - (a+2)^{5/3}}}{{u - (a+2)}}$.
Applying the formula with $n = 5/3$ and the variable $u$ approaching $a+2$,we get:
$\frac{5}{3}(a+2)^{(5/3) - 1} = \frac{5}{3}(a+2)^{2/3}$.

(C) To find the limits at $x = 3$,we evaluate the left-hand limit and right-hand limit separately.
For the right-hand limit $(x \to 3^+)$,we use the definition $f(x) = 5 - x$:
$\lim_{x \to 3^+} f(x) = 5 - 3 = 2$.
For the left-hand limit $(x \to 3^-)$,we use the definition $f(x) = \frac{2}{5-x}$:
$\lim_{x \to 3^-} f(x) = \frac{2}{5 - 3} = \frac{2}{2} = 1$.
Since $2 \neq 1$,we have $\lim_{x \to 3^+} f(x) \neq \lim_{x \to 3^-} f(x)$.

(B) We are given the limit: $\mathop {\lim }\limits_{\theta \to \pi /6} \frac{{\cot^2 \theta - 3}}{{\csc \theta - 2}}$.
Using the identity $\cot^2 \theta = \csc^2 \theta - 1$,the expression becomes:
$\mathop {\lim }\limits_{\theta \to \pi /6} \frac{{\csc^2 \theta - 1 - 3}}{{\csc \theta - 2}} = \mathop {\lim }\limits_{\theta \to \pi /6} \frac{{\csc^2 \theta - 4}}{{\csc \theta - 2}}$.
Factoring the numerator as a difference of squares:
$\mathop {\lim }\limits_{\theta \to \pi /6} \frac{{(\csc \theta - 2)(\csc \theta + 2)}}{{\csc \theta - 2}}$.
Canceling the common term $(\csc \theta - 2)$:
$\mathop {\lim }\limits_{\theta \to \pi /6} (\csc \theta + 2) = \csc(\pi /6) + 2$.
Since $\csc(\pi /6) = 2$,we have $2 + 2 = 4$.

(C) Using the standard limit formula $\mathop {\lim }\limits_{u \to a} \frac{u^n - a^n}{u - a} = n a^{n-1}$:
Let $u = 1 + x$. As $x \to 0$,$u \to 1$.
The expression becomes $\mathop {\lim }\limits_{u \to 1} \frac{u^5 - 1^5}{u^3 - 1^3}$.
Dividing numerator and denominator by $(u - 1)$:
$\mathop {\lim }\limits_{u \to 1} \frac{(u^5 - 1^5)/(u - 1)}{(u^3 - 1^3)/(u - 1)} = \frac{5(1)^{5-1}}{3(1)^{3-1}} = \frac{5}{3}$.
Alternatively,applying $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}((1+x)^5 - 1)}{\frac{d}{dx}((1+x)^3 - 1)} = \mathop {\lim }\limits_{x \to 0} \frac{5(1+x)^4}{3(1+x)^2} = \frac{5(1)^4}{3(1)^2} = \frac{5}{3}$.

(A) Given the limit: $\mathop {\lim }\limits_{x \to a} \frac{{{x^9} + {a^9}}}{{x + a}} = 9$
Since the expression is continuous at $x = a$,we can substitute $x = a$ directly:
$\frac{{{a^9} + {a^9}}}{{a + a}} = 9$
$\frac{{2{a^9}}}{{2a}} = 9$
${a^8} = 9$
$a = \pm 9^{1/8}$

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to 0^ + } \frac{x e^{1/x}}{1 + e^{1/x}}$,we divide the numerator and the denominator by $e^{1/x}$:
$\mathop {\lim }\limits_{x \to 0^ + } \frac{x}{\frac{1}{e^{1/x}} + 1} = \mathop {\lim }\limits_{x \to 0^ + } \frac{x}{e^{-1/x} + 1}$
As $x \to 0^ +$,the term $1/x \to \infty$,so $e^{-1/x} \to 0$.
Substituting these values,we get $\frac{0}{0 + 1} = 0$.

(C) To find the limit $\mathop {\lim }\limits_{x \to 1} [x]$,we evaluate the left-hand limit and the right-hand limit.
Left-hand limit: $\mathop {\lim }\limits_{x \to 1^-} [x] = \mathop {\lim }\limits_{h \to 0} [1 - h] = 0$.
Right-hand limit: $\mathop {\lim }\limits_{x \to 1^+} [x] = \mathop {\lim }\limits_{h \to 0} [1 + h] = 1$.
Since the left-hand limit $(0)$ is not equal to the right-hand limit $(1)$,the limit does not exist.

(B) To find the limit $\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + bx + 4}}{{{x^2} + ax + 5}}} \right)$,divide the numerator and the denominator by $x^2$:
$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{1 + \frac{b}{x} + \frac{4}{{{x^2}}}}}{{1 + \frac{a}{x} + \frac{5}{{{x^2}}}}}} \right)$
As $x \to \infty$,the terms $\frac{b}{x}$,$\frac{4}{x^2}$,$\frac{a}{x}$,and $\frac{5}{x^2}$ all approach $0$.
Therefore,the limit is $\frac{1 + 0 + 0}{1 + 0 + 0} = 1$.

(B) We are given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{x}$.
To solve this,we add and subtract $1$ in the numerator:
$\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1 - ({b^x} - 1)}}{x} = \mathop {\lim }\limits_{x \to 0} \left( \frac{{{a^x} - 1}}{x} - \frac{{{b^x} - 1}}{x} \right)$.
Using the standard limit formula $\mathop {\lim }\limits_{x \to 0} \frac{{{k^x} - 1}}{x} = \log k$:
$= \log a - \log b$.
Using the logarithmic property $\log m - \log n = \log \left( \frac{m}{n} \right)$:
$= \log \left( \frac{a}{b} \right)$.

(C) We know that the sum of the squares of the first $n$ natural numbers is given by $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
Substituting this into the limit expression:
$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{n(n + 1)(2n + 1)}}{{6{n^3}}}} \right]$
$= \mathop {\lim }\limits_{n \to \infty } \frac{n}{n} \cdot \frac{(n + 1)}{n} \cdot \frac{(2n + 1)}{n} \cdot \frac{1}{6}$
$= \mathop {\lim }\limits_{n \to \infty } \left( 1 \right) \left( 1 + \frac{1}{n} \right) \left( 2 + \frac{1}{n} \right) \cdot \frac{1}{6}$
$= 1 \cdot (1 + 0) \cdot (2 + 0) \cdot \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.
Thus,the correct option is $C$.

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} + {a^2}} - \sqrt {{x^2} + {b^2}} }}{{\sqrt {{x^2} + {c^2}} - \sqrt {{x^2} + {d^2}} }}$,we rationalize the numerator and the denominator:
Multiply the numerator and denominator by their respective conjugates:
$= \mathop {\lim }\limits_{x \to \infty } \frac{(\sqrt {x^2 + a^2} - \sqrt {x^2 + b^2})(\sqrt {x^2 + a^2} + \sqrt {x^2 + b^2})}{(\sqrt {x^2 + c^2} - \sqrt {x^2 + d^2})(\sqrt {x^2 + c^2} + \sqrt {x^2 + d^2})} \times \frac{(\sqrt {x^2 + c^2} + \sqrt {x^2 + d^2})}{(\sqrt {x^2 + a^2} + \sqrt {x^2 + b^2})} $
$= \mathop {\lim }\limits_{x \to \infty } \frac{(x^2 + a^2) - (x^2 + b^2)}{(x^2 + c^2) - (x^2 + d^2)} \times \frac{\sqrt {x^2 + c^2} + \sqrt {x^2 + d^2}}{\sqrt {x^2 + a^2} + \sqrt {x^2 + b^2}} $
$= \mathop {\lim }\limits_{x \to \infty } \frac{a^2 - b^2}{c^2 - d^2} \times \frac{x\sqrt {1 + c^2/x^2} + x\sqrt {1 + d^2/x^2}}{x\sqrt {1 + a^2/x^2} + x\sqrt {1 + b^2/x^2}} $
As $x \to \infty$,the terms $a^2/x^2, b^2/x^2, c^2/x^2, d^2/x^2$ approach $0$.
$= \frac{a^2 - b^2}{c^2 - d^2} \times \frac{1 + 1}{1 + 1} = \frac{a^2 - b^2}{c^2 - d^2}$.

(A) To evaluate the limit $\mathop {\lim }\limits_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}$,we rationalize the denominator:
$\mathop {\lim }\limits_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{(\sqrt{x - 2} - \sqrt{4 - x})(\sqrt{x - 2} + \sqrt{4 - x})}$
$= \mathop {\lim }\limits_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{(x - 2) - (4 - x)}$
$= \mathop {\lim }\limits_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{2x - 6}$
$= \mathop {\lim }\limits_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{2(x - 3)}$
$= \mathop {\lim }\limits_{x \to 3} \frac{\sqrt{x - 2} + \sqrt{4 - x}}{2}$
$= \frac{\sqrt{3 - 2} + \sqrt{4 - 3}}{2} = \frac{1 + 1}{2} = 1$.
Alternatively,applying $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 3} \frac{\frac{d}{dx}(x - 3)}{\frac{d}{dx}(\sqrt{x - 2} - \sqrt{4 - x})} = \mathop {\lim }\limits_{x \to 3} \frac{1}{\frac{1}{2\sqrt{x - 2}} - \frac{-1}{2\sqrt{4 - x}}} = \frac{1}{\frac{1}{2} + \frac{1}{2}} = 1$.

(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to \infty } \frac{{(x - 1)(2x + 3)}}{{{x^2}}}$.
First,expand the numerator: $(x - 1)(2x + 3) = 2x^2 + 3x - 2x - 3 = 2x^2 + x - 3$.
Now,substitute this into the limit expression: $\mathop {\lim }\limits_{x \to \infty } \frac{{2x^2 + x - 3}}{{x^2}}$.
Divide each term in the numerator by $x^2$: $\mathop {\lim }\limits_{x \to \infty } (\frac{{2x^2}}{{x^2}} + \frac{x}{{x^2}} - \frac{3}{{x^2}}) = \mathop {\lim }\limits_{x \to \infty } (2 + \frac{1}{x} - \frac{3}{{x^2}})$.
As $x \to \infty$,$\frac{1}{x} \to 0$ and $\frac{3}{{x^2}} \to 0$.
Therefore,the limit is $2 + 0 - 0 = 2$.

(C) The sum of the cubes of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}$.
Substituting this into the limit expression:
$\mathop {\lim }\limits_{n \to \infty } \frac{n^2(n+1)^2}{4n^4} = \mathop {\lim }\limits_{n \to \infty } \frac{n^2(n^2 + 2n + 1)}{4n^4}$.
$= \mathop {\lim }\limits_{n \to \infty } \frac{n^4 + 2n^3 + n^2}{4n^4}$.
Divide numerator and denominator by $n^4$:
$= \mathop {\lim }\limits_{n \to \infty } \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{4} = \frac{1 + 0 + 0}{4} = \frac{1}{4}$.

(C) Given the expression $\mathop {\lim }\limits_{x \to 0} \frac{{{y^2}}}{x}$ where ${y^2} = ax + b{x^2} + c{x^3}$.
Substituting the value of ${y^2}$ in the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{{ax + b{x^2} + c{x^3}}}{x}$
Divide each term in the numerator by $x$:
$\mathop {\lim }\limits_{x \to 0} (a + bx + c{x^2})$
Now,apply the limit $x \to 0$:
$a + b(0) + c(0)^2 = a$.

(B) To evaluate the limit $\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{{x^2} + 5x - 6}}$,we first check the form by substituting $x = 1$.
Substituting $x = 1$ gives $\frac{1^3 - 1}{1^2 + 5(1) - 6} = \frac{0}{0}$,which is an indeterminate form.
We factorize the numerator and the denominator:
Numerator: $x^3 - 1 = (x - 1)(x^2 + x + 1)$
Denominator: $x^2 + 5x - 6 = (x - 1)(x + 6)$
Now,the limit becomes $\mathop {\lim }\limits_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 6)}$.
Canceling the common factor $(x - 1)$,we get $\mathop {\lim }\limits_{x \to 1} \frac{x^2 + x + 1}{x + 6}$.
Substituting $x = 1$ gives $\frac{1^2 + 1 + 1}{1 + 6} = \frac{3}{7}$.

(A) Given limit: $L = \mathop {\lim }\limits_{x \to a} \frac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }}$
Rationalizing the numerator and denominator:
$L = \mathop {\lim }\limits_{x \to a} \frac{(\sqrt {a + 2x} - \sqrt {3x})(\sqrt {a + 2x} + \sqrt {3x})}{(\sqrt {3a + x} - 2\sqrt x)(\sqrt {3a + x} + 2\sqrt x)} \times \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$
$L = \mathop {\lim }\limits_{x \to a} \frac{(a + 2x - 3x)}{(3a + x - 4x)} \times \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$
$L = \mathop {\lim }\limits_{x \to a} \frac{a - x}{3a - 3x} \times \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$
$L = \mathop {\lim }\limits_{x \to a} \frac{a - x}{3(a - x)} \times \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$
$L = \frac{1}{3} \times \frac{\sqrt {3a + a} + 2\sqrt a}{\sqrt {a + 2a} + \sqrt {3a}} = \frac{1}{3} \times \frac{2\sqrt {4a} + 2\sqrt a}{2\sqrt {3a}} = \frac{1}{3} \times \frac{2(2\sqrt a) + 2\sqrt a}{2\sqrt {3a}} = \frac{1}{3} \times \frac{6\sqrt a}{2\sqrt {3a}} = \frac{1}{3} \times \frac{3}{\sqrt 3} = \frac{1}{\sqrt 3} = \frac{\sqrt 3}{3}$.
Wait,re-evaluating the limit: $\frac{1}{3} \times \frac{2\sqrt{4a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \times \frac{4\sqrt{a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \times \frac{6\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \times \frac{3}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Correction: The provided option $B$ is $\frac{2}{3\sqrt{3}}$. Let's re-check the simplification: $\frac{1}{3} \times \frac{2\sqrt{4a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \times \frac{4\sqrt{a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{6\sqrt{a}}{6\sqrt{3a}} = \frac{1}{\sqrt{3}}$.
Actually,$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$. The correct value is $\frac{1}{\sqrt{3}}$.

(B) We are given the limit: $\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}}$.
Notice that the denominator can be factored as a difference of squares: $1 - {x^{ - 2/3}} = (1 - {x^{ - 1/3}})(1 + {x^{ - 1/3}})$.
Substituting this into the limit expression:
$\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^{ - 1/3}}}}{{(1 - {x^{ - 1/3}})(1 + {x^{ - 1/3}})}}$.
Canceling the common term $(1 - {x^{ - 1/3}})$ (since $x \to 1$,$x \neq 1$):
$\mathop {\lim }\limits_{x \to 1} \frac{1}{{1 + {x^{ - 1/3}}}} = \frac{1}{{1 + 1^{ - 1/3}}} = \frac{1}{{1 + 1}} = \frac{1}{2}$.
Alternatively,applying $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 1} \frac{\frac{d}{dx}(1 - x^{-1/3})}{\frac{d}{dx}(1 - x^{-2/3})} = \mathop {\lim }\limits_{x \to 1} \frac{\frac{1}{3}x^{-4/3}}{\frac{2}{3}x^{-5/3}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{2}x^{1/3} = \frac{1}{2}$.

(B) Given limit: $L = \mathop {\lim }\limits_{x \to 0} \frac{{x({2^x} - 1)}}{{1 - \cos x}}$
Using the standard limits $\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x} = \log 2$ and $1 - \cos x = 2\sin^2(\frac{x}{2})$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x({2^x} - 1)}}{{2\sin^2(\frac{x}{2})}}$
$L = \mathop {\lim }\limits_{x \to 0} \left( \frac{{{2^x} - 1}}{x} \right) \cdot \left( \frac{x^2}{2\sin^2(\frac{x}{2})} \right)$
$L = (\log 2) \cdot \mathop {\lim }\limits_{x \to 0} \frac{1}{2 \left( \frac{\sin(x/2)}{x} \right)^2}$
Since $\mathop {\lim }\limits_{x \to 0} \frac{\sin(x/2)}{x/2} = 1$,we have $\mathop {\lim }\limits_{x \to 0} \frac{\sin(x/2)}{x} = \frac{1}{2}$:
$L = (\log 2) \cdot \frac{1}{2 \cdot (1/2)^2} = (\log 2) \cdot \frac{1}{2 \cdot (1/4)} = (\log 2) \cdot 2 = 2\log 2 = \log 4$.

(A) We have $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{x^3}}}$.
Using the expansion $\tan x = x + \frac{x^3}{3} + \dots$ and $\sin x = x - \frac{x^3}{6} + \dots$,we get:
$\mathop {\lim }\limits_{x \to 0} \frac{{(x + \frac{x^3}{3} + \dots) - (x - \frac{x^3}{6} + \dots)}}{{{x^3}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{x^3}{3} + \frac{x^3}{6}}}{{{x^3}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{3x^3}{6}}}{{{x^3}}} = \frac{1}{2}$.

(C) Using the trigonometric identity $\sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$:
Let $C = 2+x$ and $D = 2-x$.
Then $\sin(2+x) - \sin(2-x) = 2 \cos \left( \frac{2+x+2-x}{2} \right) \sin \left( \frac{2+x-(2-x)}{2} \right) = 2 \cos(2) \sin(x)$.
Substituting this into the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{2 \cos 2 \sin x}{x} = 2 \cos 2 \cdot \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}$.
Since $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,the result is $2 \cos 2 \cdot 1 = 2 \cos 2$.

(B) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 3x + 1}}{{{x^2} - 1}}$,we divide the numerator and the denominator by the highest power of $x$,which is ${x^2}$.
$\mathop {\lim }\limits_{x \to \infty } \frac{{2 - (3/x) + (1/{x^2})}}{{1 - (1/{x^2})}}$
As $x \to \infty$,the terms $(3/x)$,$(1/{x^2})$,and $(1/{x^2})$ approach $0$.
Therefore,the limit becomes $\frac{{2 - 0 + 0}}{{1 - 0}} = \frac{2}{1} = 2$.

(C) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2} + 2x - 1}}{{2{x^2} - 3x - 3}}$,divide the numerator and the denominator by the highest power of $x$,which is $x^2$.
$\mathop {\lim }\limits_{x \to \infty } \frac{{3 + \frac{2}{x} - \frac{1}{{{x^2}}}}}{{2 - \frac{3}{x} - \frac{3}{{{x^2}}}}}$
As $x \to \infty$,the terms $\frac{2}{x}$,$\frac{1}{{{x^2}}}$,$\frac{3}{x}$,and $\frac{3}{{{x^2}}}$ approach $0$.
Therefore,the limit is $\frac{{3 + 0 - 0}}{{2 - 0 - 0}} = \frac{3}{2}$.

(C) To find the limit $\mathop {\lim }\limits_{x \to 2} \frac{|x - 2|}{x - 2}$,we evaluate the left-hand limit and the right-hand limit.
Left-hand limit $(LHL)$: $\mathop {\lim }\limits_{x \to 2^-} \frac{|x - 2|}{x - 2} = \mathop {\lim }\limits_{h \to 0} \frac{|2 - h - 2|}{2 - h - 2} = \mathop {\lim }\limits_{h \to 0} \frac{|-h|}{-h} = \mathop {\lim }\limits_{h \to 0} \frac{h}{-h} = -1$.
Right-hand limit $(RHL)$: $\mathop {\lim }\limits_{x \to 2^+} \frac{|x - 2|}{x - 2} = \mathop {\lim }\limits_{h \to 0} \frac{|2 + h - 2|}{2 + h - 2} = \mathop {\lim }\limits_{h \to 0} \frac{|h|}{h} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h} = 1$.
Since the left-hand limit $(-1)$ is not equal to the right-hand limit $(1)$,the limit does not exist.

(C) To evaluate the limit $\mathop {\lim }\limits_{x \to \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}}$,we factor out the highest power of $x$ from each term:
$= \mathop {\lim }\limits_{x \to \infty } \frac{{{{[x(2 + \frac{1}{x})]}^{40}}{{[x(4 - \frac{1}{x})]}^5}}}{{{{[x(2 + \frac{3}{x})]}^{45}}}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{40}}{{(2 + \frac{1}{x})}^{40}} \cdot {x^5}{{(4 - \frac{1}{x})}^5}}}{{{x^{45}}{{(2 + \frac{3}{x})}^{45}}}}$
$= \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{45}}{{(2 + \frac{1}{x})}^{40}}{{(4 - \frac{1}{x})}^5}}}{{{x^{45}}{{(2 + \frac{3}{x})}^{45}}}}$
$= \frac{{{{(2 + 0)}^{40}}{{(4 - 0)}^5}}}{{{{(2 + 0)}^{45}}}}$
$= \frac{{{2^{40}} \cdot {4^5}}}{{{2^{45}}}}$
$= \frac{{{2^{40}} \cdot {{(2^2)}^5}}}{{{2^{45}}}}$
$= \frac{{{2^{40}} \cdot {2^{10}}}}{{{2^{45}}}}$
$= \frac{{{2^{50}}}}{{{2^{45}}}} = {2^5} = 32$

(B) Let $\tan^{-1} 2x = \theta$.
Then $2x = \tan \theta$,which implies $x = \frac{1}{2} \tan \theta$.
As $x \to 0$,$\theta \to 0$.
Substituting these into the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{x}{\tan^{-1} 2x} = \mathop {\lim }\limits_{\theta \to 0} \frac{\frac{1}{2} \tan \theta}{\theta} = \frac{1}{2} \mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta}$.
Since $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$,the result is $\frac{1}{2} \times 1 = \frac{1}{2}$.

(D) Let $\theta = \pi - 2x$. As $x \to \frac{\pi}{2}$,$\theta \to 0$.
Then $2x = \pi - \theta$,so $\cos 2x = \cos(\pi - \theta) = -\cos \theta$.
The expression becomes $\mathop {\lim }\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2}$.
Using the identity $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$,we get $\mathop {\lim }\limits_{\theta \to 0} \frac{2 \sin^2(\frac{\theta}{2})}{\theta^2}$.
This simplifies to $\mathop {\lim }\limits_{\theta \to 0} 2 \cdot \frac{\sin^2(\frac{\theta}{2})}{4(\frac{\theta}{2})^2} = 2 \cdot \frac{1}{4} = \frac{1}{2}$.

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to \pi /2} \frac{\tan 3x}{x}$.
As $x \to \pi /2$,the numerator $\tan 3x$ approaches $\tan(3\pi / 2)$,which is undefined (tends to $\pm \infty$ depending on the direction).
The denominator $x$ approaches $\pi / 2$.
Since the numerator is undefined and the denominator is a finite non-zero value,the limit does not exist in the real number system and tends to $\infty$ in the context of extended real numbers.
Therefore,the correct option is $A$.

(D) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3 + x} - \sqrt {3 - x} }}{x}$,we rationalize the numerator:
Multiply the numerator and denominator by $(\sqrt{3+x} + \sqrt{3-x})$:
$= \mathop {\lim }\limits_{x \to 0} \frac{(\sqrt{3+x} - \sqrt{3-x})(\sqrt{3+x} + \sqrt{3-x})}{x(\sqrt{3+x} + \sqrt{3-x})}$
$= \mathop {\lim }\limits_{x \to 0} \frac{(3+x) - (3-x)}{x(\sqrt{3+x} + \sqrt{3-x})}$
$= \mathop {\lim }\limits_{x \to 0} \frac{2x}{x(\sqrt{3+x} + \sqrt{3-x})}$
$= \mathop {\lim }\limits_{x \to 0} \frac{2}{\sqrt{3+x} + \sqrt{3-x}}$
$= \frac{2}{\sqrt{3+0} + \sqrt{3-0}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.

(A) Let $f(x) = \log x$. Then $f'(x) = \frac{1}{x}$.
The first limit is $\mathop {\lim }\limits_{x \to 0} \frac{{\log (a + x) - \log a}}{x} = f'(a) = \frac{1}{a}$.
The second limit is $\mathop {\lim }\limits_{x \to e} \frac{{\log x - 1}}{{x - e}}$. Since $\log e = 1$,this is $\mathop {\lim }\limits_{x \to e} \frac{{\log x - \log e}}{{x - e}} = f'(e) = \frac{1}{e}$.
Substituting these into the given equation: $\frac{1}{a} + k \cdot \frac{1}{e} = 1$.
$\frac{k}{e} = 1 - \frac{1}{a} = \frac{a - 1}{a}$.
Therefore,$k = e \left( \frac{a - 1}{a} \right) = e \left( 1 - \frac{1}{a} \right)$.

(D) We know that $1 - \cos 2x = 2 \sin^2 x$.
Substituting this into the expression,we get $\sqrt{\frac{1}{2}(2 \sin^2 x)} = \sqrt{\sin^2 x} = |\sin x|$.
Thus,the limit becomes $\mathop {\lim }\limits_{x \to 0} \frac{|\sin x|}{x}$.
For the right-hand limit $(x \to 0^+)$,$|\sin x| = \sin x$,so $\mathop {\lim }\limits_{x \to 0^+} \frac{\sin x}{x} = 1$.
For the left-hand limit $(x \to 0^-)$,$|\sin x| = -\sin x$,so $\mathop {\lim }\limits_{x \to 0^-} \frac{-\sin x}{x} = -1$.
Since the left-hand limit and right-hand limit are not equal,the limit does not exist.

(D) We use the standard limit formula $\mathop {\lim }\limits_{u \to 0} \frac{e^u - 1}{u} = 1$.
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\alpha x}} - {e^{\beta x}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{\alpha x}} - 1 - ({e^{\beta x}} - 1)}}{x}$
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{{{e^{\alpha x}} - 1}}{x} - \frac{{{e^{\beta x}} - 1}}{x} \right)$
$= \mathop {\lim }\limits_{x \to 0} \left( \alpha \cdot \frac{{{e^{\alpha x}} - 1}}{{\alpha x}} - \beta \cdot \frac{{{e^{\beta x}} - 1}}{{\beta x}} \right)$
$= \alpha(1) - \beta(1) = \alpha - \beta$.