Concept of limits, Evaluation of algebric limits Questions in English

(D) To find $\lim_{x \to 0} f(x)$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$ at $x = 0$.
For the $LHL$: $\lim_{x \to 0^-} f(x)$. As $x \to 0^-$,$x$ is slightly less than $0$,so $[x] = -1$.
Thus,$f(x) = \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1} = \sin(1)$.
So,$\lim_{x \to 0^-} f(x) = \sin(1)$.
For the $RHL$: $\lim_{x \to 0^+} f(x)$. As $x \to 0^+$,$x$ is in the interval $[0, 1)$,so $[x] = 0$.
According to the definition of $f(x)$,when $[x] = 0$,$f(x) = 0$.
So,$\lim_{x \to 0^+} f(x) = 0$.
Since $\lim_{x \to 0^-} f(x) = \sin(1)$ and $\lim_{x \to 0^+} f(x) = 0$,the left-hand limit is not equal to the right-hand limit.
Therefore,$\lim_{x \to 0} f(x)$ does not exist.

(C) Given the limit: $\mathop {\lim }\limits_{n \to \infty } \frac{1 - 10^n}{1 + 10^{n+1}}$
Divide the numerator and denominator by $10^n$:
$= \mathop {\lim }\limits_{n \to \infty } \frac{\frac{1}{10^n} - 1}{\frac{1}{10^n} + 10}$
As $n \to \infty$,$\frac{1}{10^n} \to 0$:
$= \frac{0 - 1}{0 + 10} = -\frac{1}{10}$
Comparing this with the given expression $\frac{-\alpha}{10}$,we get:
$-\frac{1}{10} = -\frac{\alpha}{10}$
Therefore,$\alpha = 1$.

(B) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\log (1 + {x^3})}}{{{{\sin }^3}x}}$.
Since the form is $\frac{0}{0}$ as $x \to 0$,we can use the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{{\log (1 + u)}}{u} = 1$ where $u = x^3$.
$\mathop {\lim }\limits_{x \to 0} \frac{{\log (1 + {x^3})}}{{{{\sin }^3}x}} = \mathop {\lim }\limits_{x \to 0} \left( \frac{{\log (1 + {x^3})}}{{{x^3}}} \times \frac{{{x^3}}}{{{{\sin }^3}x}} \right)$
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{{\log (1 + {x^3})}}{{{x^3}}} \right) \times \mathop {\lim }\limits_{x \to 0} \left( \frac{x}{{\sin x}} \right)^3$
$= 1 \times (1)^3 = 1$.

(C) Given the limit $\mathop {\lim }\limits_{n \to \infty } \frac{{{x^n}}}{{{x^n} + 1}}$.
Since $x < -1$,we have $|x| > 1$,which implies that $\mathop {\lim }\limits_{n \to \infty } x^n$ does not exist in the finite sense,but we can divide the numerator and denominator by $x^n$:
$\mathop {\lim }\limits_{n \to \infty } \frac{{{x^n}}}{{{x^n}(1 + \frac{1}{x^n})}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{1 + \frac{1}{x^n}}$.
As $n \to \infty$ and $|x| > 1$,$\frac{1}{x^n} \to 0$.
Therefore,the limit is $\frac{1}{1 + 0} = 1$.

(B) The given expression is $\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1 + 2 + 3 + \dots + n}{1 - {n^2}}} \right]$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,the expression becomes:
$\mathop {\lim }\limits_{n \to \infty } \frac{n(n+1)}{2(1 - {n^2})}$
$= \mathop {\lim }\limits_{n \to \infty } \frac{n^2 + n}{2 - 2n^2}$
Dividing the numerator and denominator by $n^2$:
$= \mathop {\lim }\limits_{n \to \infty } \frac{1 + \frac{1}{n}}{\frac{2}{n^2} - 2} = \frac{1 + 0}{0 - 2} = -\frac{1}{2}$.

(C) Given limit $L = \mathop {\lim }\limits_{n \to \infty } \frac{{{1^2} + {2^2} + {3^2} + \dots + {n^2}}}{{{n^3} + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum_{k=1}^{n} k^2}}{{{n^3} + 1}}$
Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$L = \mathop {\lim }\limits_{n \to \infty } \frac{n(n+1)(2n+1)}{6(n^3+1)}$
$L = \mathop {\lim }\limits_{n \to \infty } \frac{1}{6} \cdot \frac{n^3(1 + \frac{1}{n})(2 + \frac{1}{n})}{n^3(1 + \frac{1}{n^3})}$
$L = \frac{1}{6} \cdot \frac{(1+0)(2+0)}{(1+0)} = \frac{2}{6} = \frac{1}{3}$

(A) We are given that ${S_n} = \sum\limits_{k = 1}^n {{a_k}}$.
Therefore,${S_{n + 1}} - {S_n} = {a_{n + 1}}$.
We know that $\sum\limits_{k = 1}^n k = \frac{n(n + 1)}{2}$.
Thus,the limit becomes $\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{\sqrt{\frac{n(n + 1)}{2}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}} \cdot \sqrt{2}}}{{\sqrt{n(n + 1)}}}$.
Since $\mathop {\lim }\limits_{n \to \infty } {a_n} = a$,we have $\mathop {\lim }\limits_{n \to \infty } {a_{n + 1}} = a$.
As $n \to \infty$,the numerator approaches $a \cdot \sqrt{2}$ (a finite value) and the denominator $\sqrt{n(n + 1)} \to \infty$.
Therefore,the limit is $\frac{a \cdot \sqrt{2}}{\infty} = 0$.

(B) We know the identity: $\cos A \cos 2A \cos 4A \dots \cos (2^{n-1}A) = \frac{\sin (2^n A)}{2^n \sin A}$.
Let $A = \frac{x}{2^n}$. Then the product becomes:
$\cos \left( \frac{x}{2^n} \right) \cos \left( \frac{x}{2^{n-1}} \right) \dots \cos \left( \frac{x}{2} \right) = \frac{\sin (2^n \cdot \frac{x}{2^n})}{2^n \sin (\frac{x}{2^n})} = \frac{\sin x}{2^n \sin (\frac{x}{2^n})}$.
Taking the limit as $n \to \infty$:
$\mathop {\lim }\limits_{n \to \infty } \frac{\sin x}{2^n \sin (\frac{x}{2^n})} = \sin x \cdot \mathop {\lim }\limits_{n \to \infty } \frac{1}{2^n \sin (\frac{x}{2^n})}$.
Since $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,let $\theta = \frac{x}{2^n}$. As $n \to \infty$,$\theta \to 0$:
$\mathop {\lim }\limits_{n \to \infty } \frac{x/2^n}{\sin (x/2^n)} = 1$.
Thus,the expression becomes $\frac{\sin x}{x} \cdot 1 = \frac{\sin x}{x}$.

(C) The given expression is a geometric series with first term $a = \frac{1}{2}$ and common ratio $r = \frac{1}{2}$.
The sum of the first $n$ terms of a geometric series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.
Substituting the values,we get $S_n = \frac{\frac{1}{2}(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} = \frac{\frac{1}{2}(1 - \frac{1}{2^n})}{\frac{1}{2}} = 1 - \frac{1}{2^n}$.
Now,taking the limit as $n \to \infty$:
$\mathop {\lim }\limits_{n \to \infty } (1 - \frac{1}{2^n}) = 1 - 0 = 1$.

(A) The given limit is $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}} + \frac{3}{{{n^2}}} + \dots + \frac{n}{{{n^2}}}} \right)$.
We can factor out $\frac{1}{{{n^2}}}$ from the sum:
$= \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2}}} (1 + 2 + 3 + \dots + n)$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:
$= \mathop {\lim }\limits_{n \to \infty } \frac{n(n+1)}{2{n^2}}$.
Simplify the expression:
$= \mathop {\lim }\limits_{n \to \infty } \frac{n^2 + n}{2n^2} = \mathop {\lim }\limits_{n \to \infty } \left( \frac{1}{2} + \frac{1}{2n} \right)$.
As $n \to \infty$,$\frac{1}{n} \to 0$:
$= \frac{1}{2} + 0 = \frac{1}{2}$.

(A) We know that the sum of the first $n$ natural numbers is given by $\sum n = \frac{n(n + 1)}{2}$.
Substituting this into the limit expression:
$\mathop {\lim }\limits_{n \to \infty } \frac{1 - n^2}{\frac{n(n + 1)}{2}}$
$= \mathop {\lim }\limits_{n \to \infty } \frac{2(1 - n)(1 + n)}{n(n + 1)}$
$= \mathop {\lim }\limits_{n \to \infty } \frac{2(1 - n)}{n}$
$= \mathop {\lim }\limits_{n \to \infty } 2 \left( \frac{1}{n} - 1 \right)$
As $n \to \infty$,$\frac{1}{n} \to 0$.
Therefore,the limit is $2(0 - 1) = -2$.

(B) The numerator is the sum of the series $1 - 2 + 3 - 4 + \dots - 2n$.
This can be grouped as $(1 - 2) + (3 - 4) + \dots + ((2n-1) - 2n)$.
There are $n$ such pairs,each summing to $-1$.
Thus,the numerator is $n \times (-1) = -n$.
Now,$\mathop {\lim }\limits_{n \to \infty } {x_n} = \mathop {\lim }\limits_{n \to \infty } \frac{-n}{\sqrt {n^2 + 1} + \sqrt {4n^2 - 1}}$.
Dividing the numerator and denominator by $n$:
$\mathop {\lim }\limits_{n \to \infty } \frac{-1}{\sqrt {1 + \frac{1}{n^2}} + \sqrt {4 - \frac{1}{n^2}}} = \frac{-1}{\sqrt{1+0} + \sqrt{4-0}} = \frac{-1}{1+2} = -\frac{1}{3}$.

(D) We are given the limit: $\mathop {\lim }\limits_{x \to \infty } \frac{{{{(x + 1)}^{10}} + {{(x + 2)}^{10}} + \dots + {{(x + 100)}^{10}}}}{{{x^{10}} + {{10}^{10}}}}$
Divide the numerator and denominator by $x^{10}$:
$= \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {1 + \frac{1}{x}} \right)}^{10}} + {{\left( {1 + \frac{2}{x}} \right)}^{10}} + \dots + {{\left( {1 + \frac{{100}}{x}} \right)}^{10}}}}{{1 + \frac{{{{10}^{10}}}}{{{x^{10}}}}}}$
As $x \to \infty$,the terms $\frac{k}{x} \to 0$ for any constant $k$,and $\frac{10^{10}}{x^{10}} \to 0$.
Therefore,the expression becomes:
$= \frac{{{{(1 + 0)}^{10}} + {{(1 + 0)}^{10}} + \dots + {{(1 + 0)}^{10}}}}{{1 + 0}}$
Since there are $100$ terms in the numerator,each equal to $1^{10} = 1$:
$= \frac{1 + 1 + \dots + 1 \text{ (100 times)}}{1} = 100$.

(B) We know that the sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$.
Substituting this into the limit,we get:
$\mathop {\lim }\limits_{n \to \infty } \frac{n(n + 1)}{2({n^2} + 100)}$
$= \mathop {\lim }\limits_{n \to \infty } \frac{{n^2} + n}{2{n^2} + 200}$
Dividing the numerator and the denominator by ${n^2}$:
$= \mathop {\lim }\limits_{n \to \infty } \frac{1 + \frac{1}{n}}{2 + \frac{200}{n^2}}$
As $n \to \infty$,$\frac{1}{n} \to 0$ and $\frac{200}{n^2} \to 0$.
Therefore,the limit is $\frac{1 + 0}{2 + 0} = \frac{1}{2}$.

(B) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$.
Rationalizing the numerator and denominator:
$L = \mathop {\lim }\limits_{x \to 0} \frac{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})}{(\sqrt{2+3x} - \sqrt{2-3x})(\sqrt{2+3x} + \sqrt{2-3x})} \times \frac{\sqrt{2+3x} + \sqrt{2-3x}}{\sqrt{1+x} + \sqrt{1-x}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{(1+x) - (1-x)}{(2+3x) - (2-3x)} \times \frac{\sqrt{2+3x} + \sqrt{2-3x}}{\sqrt{1+x} + \sqrt{1-x}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{2x}{6x} \times \frac{\sqrt{2+3x} + \sqrt{2-3x}}{\sqrt{1+x} + \sqrt{1-x}}$
$L = \frac{1}{3} \times \frac{\sqrt{2} + \sqrt{2}}{1 + 1} = \frac{1}{3} \times \frac{2\sqrt{2}}{2} = \frac{\sqrt{2}}{3}$.
Since $\sqrt{2} \approx 1.414$,then $\frac{\sqrt{2}}{3} \approx \frac{1.414}{3} \approx 0.471$.
Thus,$0 < \frac{\sqrt{2}}{3} < \frac{1}{2}$.
Therefore,the correct option is $B$.

(C) Let $f(x) = 1 - x + [x - 1] + [1 - x]$.
We evaluate the left-hand limit as $x \to 1^-$:
$\mathop {\lim }\limits_{x \to 1^-} f(x) = \mathop {\lim }\limits_{h \to 0^+} (1 - (1 - h) + [1 - h - 1] + [1 - (1 - h)])$
$= \mathop {\lim }\limits_{h \to 0^+} (h + [-h] + [h]) = \mathop {\lim }\limits_{h \to 0^+} (h - 1 + 0) = -1$.
We evaluate the right-hand limit as $x \to 1^+$:
$\mathop {\lim }\limits_{x \to 1^+} f(x) = \mathop {\lim }\limits_{h \to 0^+} (1 - (1 + h) + [1 + h - 1] + [1 - (1 + h)])$
$= \mathop {\lim }\limits_{h \to 0^+} (-h + [h] + [-h]) = \mathop {\lim }\limits_{h \to 0^+} (-h + 0 - 1) = -1$.
Since the left-hand limit equals the right-hand limit,the limit exists and is equal to $-1$.

(B) Given the sequence $x_1 = 3$ and $x_{n+1} = \sqrt{2 + x_n}$.
We calculate the first few terms:
$x_2 = \sqrt{2 + x_1} = \sqrt{2 + 3} = \sqrt{5} \approx 2.236$
$x_3 = \sqrt{2 + x_2} = \sqrt{2 + \sqrt{5}} \approx \sqrt{4.236} \approx 2.058$
It is observed that $x_1 > x_2 > x_3 > \dots$. The sequence is monotonically decreasing and bounded below by $2$.
Since the sequence is monotonic and bounded,it must converge to a limit $L$.
Let $\lim_{n \to \infty} x_n = L$.
Taking the limit on both sides of the recurrence relation $x_{n+1} = \sqrt{2 + x_n}$:
$L = \sqrt{2 + L}$
$L^2 = 2 + L$
$L^2 - L - 2 = 0$
$(L - 2)(L + 1) = 0$
Since $x_n > 0$ for all $n$,the limit $L$ must be positive.
Therefore,$L = 2$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{(\cos x - 1)(\cos x - e^x)}{x^n}$.
Using Taylor series expansions near $x = 0$:
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
$e^x = 1 + x + \frac{x^2}{2!} + \dots$
Substituting these into the expression:
$\cos x - 1 = -\frac{x^2}{2} + O(x^4)$
$\cos x - e^x = (1 - \frac{x^2}{2} + \dots) - (1 + x + \frac{x^2}{2} + \dots) = -x - x^2 + O(x^3)$
Thus,the numerator is $(- \frac{x^2}{2} + O(x^4))(-x - x^2 + O(x^3)) = \frac{x^3}{2} + O(x^4)$.
For the limit to be a finite non-zero number,the power of $x$ in the denominator must match the lowest power of $x$ in the numerator.
Therefore,$n = 3$.

(C) Given $f(x) = \begin{cases} x^2 - 3, & 2 < x < 3 \\ 2x + 5, & 3 < x < 4 \end{cases}$.
First,we find the left-hand limit as $x \to 3^-$:
$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x^2 - 3) = 3^2 - 3 = 9 - 3 = 6$.
Next,we find the right-hand limit as $x \to 3^+$:
$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x + 5) = 2(3) + 5 = 6 + 5 = 11$.
The roots of the required quadratic equation are $6$ and $11$.
The equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Sum of roots $= 6 + 11 = 17$.
Product of roots $= 6 \times 11 = 66$.
Therefore,the equation is $x^2 - 17x + 66 = 0$.

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}}$
Using the identity $1 - \cos 2x = 2\sin^2 x$,the expression becomes:
$\mathop {\lim }\limits_{x \to 0} \frac{{2\sin^2 x(3 + \cos x)}}{{x \tan 4x}}$
Multiply and divide by $x$ and $4x$ to use standard limits $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$:
$= \mathop {\lim }\limits_{x \to 0} \left( 2 \cdot \left( \frac{\sin x}{x} \right)^2 \cdot \frac{x^2}{x \cdot \frac{\tan 4x}{4x} \cdot 4x} \cdot (3 + \cos x) \right)$
$= \mathop {\lim }\limits_{x \to 0} \left( 2 \cdot \left( \frac{\sin x}{x} \right)^2 \cdot \frac{1}{4 \cdot \frac{\tan 4x}{4x}} \cdot (3 + \cos x) \right)$
Substituting $x = 0$:
$= 2 \cdot (1)^2 \cdot \frac{1}{4 \cdot 1} \cdot (3 + \cos 0)$
$= 2 \cdot \frac{1}{4} \cdot (3 + 1) = 2 \cdot \frac{1}{4} \cdot 4 = 2$

(A) Given $p = \mathop {\lim }\limits_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}}$.
This is of the form $1^\infty$,so we use the formula $\mathop {\lim }\limits_{x \to a} f(x)^{g(x)} = e^{\mathop {\lim }\limits_{x \to a} (f(x)-1)g(x)}$.
$p = e^{\mathop {\lim }\limits_{x \to 0^+} (\tan^2 \sqrt{x}) \cdot \frac{1}{2x}}$.
Since $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$,we have $\mathop {\lim }\limits_{x \to 0^+} \frac{\tan^2 \sqrt{x}}{x} = \mathop {\lim }\limits_{x \to 0^+} \left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right)^2 = 1^2 = 1$.
Thus,$p = e^{\frac{1}{2} \cdot 1} = e^{1/2}$.
Taking the natural logarithm,$\log p = \log(e^{1/2}) = \frac{1}{2}$.

(B) We know that $[t] = t - \{t\}$,where $\{t\}$ is the fractional part of $t$.
Thus,the expression becomes $\lim_{x \to 0^+} x \sum_{r=1}^{15} [\frac{r}{x}] = \lim_{x \to 0^+} x \sum_{r=1}^{15} (\frac{r}{x} - \{\frac{r}{x}\})$.
$= \lim_{x \to 0^+} (\sum_{r=1}^{15} r - x \sum_{r=1}^{15} \{\frac{r}{x}\})$.
Since $0 \le \{\frac{r}{x}\} < 1$,we have $0 \le x \{\frac{r}{x}\} < x$.
As $x \to 0^+$,by the Squeeze Theorem,$x \{\frac{r}{x}\} \to 0$.
Therefore,the limit is $\sum_{r=1}^{15} r = \frac{15 \times 16}{2} = 120$.

(A) We are given the limit $L = \mathop {\lim }\limits_{n \to \infty } n \cos \left( \frac{\pi }{4n} \right) \sin \left( \frac{\pi }{4n} \right)$.
Multiply and divide by $2$ to use the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$:
$L = \mathop {\lim }\limits_{n \to \infty } \frac{n}{2} \left( 2 \sin \left( \frac{\pi }{4n} \right) \cos \left( \frac{\pi }{4n} \right) \right)$
$L = \frac{1}{2} \mathop {\lim }\limits_{n \to \infty } n \sin \left( \frac{\pi }{2n} \right)$
Let $x = \frac{\pi }{2n}$. As $n \to \infty$,$x \to 0$. Also,$n = \frac{\pi }{2x}$.
$L = \frac{1}{2} \mathop {\lim }\limits_{x \to 0 } \left( \frac{\pi }{2x} \right) \sin(x)$
$L = \frac{\pi }{4} \mathop {\lim }\limits_{x \to 0 } \frac{\sin x}{x}$
Since $\mathop {\lim }\limits_{x \to 0 } \frac{\sin x}{x} = 1$,we get $L = \frac{\pi }{4} \times 1 = \frac{\pi }{4}$.
Thus,$k = \frac{\pi }{4}$.

(D) Let $f(x) = \frac{\sqrt{1 - \cos 2(x - 1)}}{x - 1}$.
We know that $1 - \cos 2\theta = 2\sin^2 \theta$,so $1 - \cos 2(x - 1) = 2\sin^2(x - 1)$.
Thus,$f(x) = \frac{\sqrt{2\sin^2(x - 1)}}{x - 1} = \frac{\sqrt{2}|\sin(x - 1)|}{x - 1}$.
Right Hand Limit $(RHL)$: $\mathop {\lim }\limits_{x \to 1^+} f(x) = \mathop {\lim }\limits_{h \to 0^+} \frac{\sqrt{2}|\sin h|}{h} = \mathop {\lim }\limits_{h \to 0^+} \sqrt{2} \frac{\sin h}{h} = \sqrt{2}$.
Left Hand Limit $(LHL)$: $\mathop {\lim }\limits_{x \to 1^-} f(x) = \mathop {\lim }\limits_{h \to 0^+} \frac{\sqrt{2}|\sin(-h)|}{-h} = \mathop {\lim }\limits_{h \to 0^+} \frac{\sqrt{2}\sin h}{-h} = -\sqrt{2}$.
Since $\text{RHL} \neq \text{LHL}$,the limit does not exist.

(A) 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 the form is $1^\infty$.
Here,$\mathop {\lim }\limits_{x \to \infty } \frac{x^2 + 5x + 3}{x^2 + x + 3} = 1$.
So,the expression is of the form $1^\infty$.
Let $f(x) = \frac{x^2 + 5x + 3}{x^2 + x + 3} - 1 = \frac{x^2 + 5x + 3 - x^2 - x - 3}{x^2 + x + 3} = \frac{4x}{x^2 + x + 3}$.
Then,$\mathop {\lim }\limits_{x \to \infty } f(x)g(x) = \mathop {\lim }\limits_{x \to \infty } \left( \frac{4x}{x^2 + x + 3} \cdot x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{4x^2}{x^2 + x + 3}$.
Dividing numerator and denominator by $x^2$,we get $\mathop {\lim }\limits_{x \to \infty } \frac{4}{1 + \frac{1}{x} + \frac{3}{x^2}} = \frac{4}{1 + 0 + 0} = 4$.
Therefore,the limit is $e^4$.

(C) We know that $\frac{1 - \tan(x/2)}{1 + \tan(x/2)} = \tan(\frac{\pi}{4} - \frac{x}{2})$.
So,the limit becomes $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right)\,(1 - \sin x)}}{{{{(\pi - 2x)}^3}}}$.
Let $x = \frac{\pi }{2} + y$,where $y \to 0$. Then $\pi - 2x = \pi - 2(\frac{\pi}{2} + y) = -2y$.
Also,$1 - \sin x = 1 - \sin(\frac{\pi}{2} + y) = 1 - \cos y = 2\sin^2(\frac{y}{2})$.
Substituting these,we get $\mathop {\lim }\limits_{y \to 0} \frac{{\tan \left( {\frac{\pi}{4} - (\frac{\pi}{4} + \frac{y}{2})} \right)\,(2\sin^2 \frac{y}{2})}}{{{{( - 2y)}^3}}}$.
$= \mathop {\lim }\limits_{y \to 0} \frac{{-\tan(\frac{y}{2}) \cdot 2\sin^2(\frac{y}{2})}}{-8y^3} = \mathop {\lim }\limits_{y \to 0} \frac{1}{4} \cdot \frac{\tan(y/2)}{y} \cdot \left(\frac{\sin(y/2)}{y}\right)^2$.
$= \frac{1}{4} \cdot \frac{1}{2} \cdot (\frac{1}{2})^2 = \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{32}$.

(C) Let $L = \mathop {Lim}\limits_{x \to 0} \frac{{\log _{{{\sin }^2}x}}\cos x}{{\log _{{{\sin }^2}\frac{x}{2}}}\cos \frac{x}{2}}$.
Using the base change formula $\log_a b = \frac{\ln b}{\ln a}$,we get:
$L = \mathop {Lim}\limits_{x \to 0} \frac{\ln(\cos x) / \ln(\sin^2 x)}{\ln(\cos(x/2)) / \ln(\sin^2(x/2))}$.
As $x \to 0$,$\cos x \approx 1 - x^2/2$ and $\sin x \approx x$,so $\ln(\cos x) \approx -x^2/2$ and $\ln(\sin^2 x) \approx \ln(x^2) = 2\ln x$.
Similarly,$\ln(\cos(x/2)) \approx -(x/2)^2/2 = -x^2/8$ and $\ln(\sin^2(x/2)) \approx \ln(x^2/4) = 2\ln(x/2) = 2(\ln x - \ln 2) \approx 2\ln x$.
Substituting these approximations:
$L = \mathop {Lim}\limits_{x \to 0} \frac{-x^2/2}{2\ln x} \cdot \frac{2\ln x}{-x^2/8} = \frac{-x^2/2}{-x^2/8} = \frac{8}{2} = 4$.

(B) Let $L = \mathop {Limit}\limits_{x \to \infty } \frac{{2^{\frac{x^n}{e^x}}} - {3^{\frac{x^n}{e^x}}}}{x^n}$.
Since $\mathop {Limit}\limits_{x \to \infty } \frac{x^n}{e^x} = 0$ for any $n \in N$,we can substitute $u = \frac{x^n}{e^x}$,where $u \to 0$ as $x \to \infty$.
Then $L = \mathop {Limit}\limits_{u \to 0} \frac{2^u - 3^u}{x^n}$.
Since $x^n = u \cdot e^x$,we have $L = \mathop {Limit}\limits_{x \to \infty } \frac{2^u - 3^u}{u \cdot e^x} = \mathop {Limit}\limits_{u \to 0} \left( \frac{2^u - 3^u}{u} \right) \cdot \mathop {Limit}\limits_{x \to \infty } \frac{1}{e^x}$.
Using the standard limit $\mathop {Limit}\limits_{u \to 0} \frac{a^u - 1}{u} = \ln a$,we get $\mathop {Limit}\limits_{u \to 0} \frac{2^u - 3^u}{u} = \ln 2 - \ln 3 = \ln \left( \frac{2}{3} \right)$.
As $x \to \infty$,$e^x \to \infty$,so $\frac{1}{e^x} \to 0$.
Therefore,$L = \ln \left( \frac{2}{3} \right) \cdot 0 = 0$.

(C) Since $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we can write $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.
We need to evaluate $L = \lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$.
Using the identity $1 - \cos(\theta) = 2\sin^2(\frac{\theta}{2})$,we get:
$L = \lim_{x \to \alpha} \frac{2\sin^2(\frac{a(x - \alpha)(x - \beta)}{2})}{(x - \alpha)^2}$.
Multiplying and dividing by $(\frac{a(x - \beta)}{2})^2$,we have:
$L = \lim_{x \to \alpha} 2 \cdot \left[ \frac{\sin(\frac{a(x - \alpha)(x - \beta)}{2})}{\frac{a(x - \alpha)(x - \beta)}{2}} \right]^2 \cdot \frac{a^2(x - \alpha)^2(x - \beta)^2}{4(x - \alpha)^2}$.
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,the limit becomes:
$L = 2 \cdot (1)^2 \cdot \frac{a^2(\alpha - \beta)^2}{4} = \frac{a^2}{2}(\alpha - \beta)^2$.

(B) Let $f(x) = \frac{4}{{{x^2} - \frac{1}{x}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x}}{{{x^3} - 1}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x + 1 - 3x + {x^2}}}{{{x^3} - 1}} = \frac{{{x^2} + x + 1}}{{(x - 1)({x^2} + x + 1)}} = \frac{1}{{x - 1}}$.
Then,the first term is $[f(x)]^{-1} = x - 1$.
The second term is $\frac{{3({x^4} - 1)}}{{{x^3} - \frac{1}{x}}} = \frac{{3x({x^4} - 1)}}{{{x^4} - 1}} = 3x$.
Thus,the expression becomes $\mathop {\lim}\limits_{x \to 1} [(x - 1) + 3x] = \mathop {\lim}\limits_{x \to 1} (4x - 1) = 4(1) - 1 = 3$.

(C) Let $M$ be the midpoint of $BC$. In the right triangle $ABM$,$AM = h$. Let $BM = x$. By the property of chords,$x^2 = h(2r - h)$,so $x = \sqrt{2rh - h^2}$.
$AB = AC = \sqrt{h^2 + x^2} = \sqrt{h^2 + 2rh - h^2} = \sqrt{2rh}$.
Perimeter $P = AB + AC + BC = 2\sqrt{2rh} + 2\sqrt{2rh - h^2}$.
Area $\Delta = \frac{1}{2} \times BC \times AM = \frac{1}{2} \times (2\sqrt{2rh - h^2}) \times h = h\sqrt{2rh - h^2}$.
We want to evaluate $\mathop {\lim }\limits_{h \to 0} \frac{\Delta }{{{P^3}}} = \mathop {\lim }\limits_{h \to 0} \frac{h\sqrt{2rh - h^2}}{[2\sqrt{2rh} + 2\sqrt{2rh - h^2}]^3}$.
$= \mathop {\lim }\limits_{h \to 0} \frac{h\sqrt{h}\sqrt{2r - h}}{8[\sqrt{2rh} + \sqrt{2rh - h^2}]^3} = \mathop {\lim }\limits_{h \to 0} \frac{h^{3/2}\sqrt{2r - h}}{8(h^{1/2})^3[\sqrt{2r} + \sqrt{2r - h}]^3}$.
$= \frac{\sqrt{2r}}{8[\sqrt{2r} + \sqrt{2r}]^3} = \frac{\sqrt{2r}}{8[2\sqrt{2r}]^3} = \frac{\sqrt{2r}}{8 \times 8 \times 2r \times \sqrt{2r}} = \frac{1}{128r}$.

(A) We know that $\sin(3x) = 3\sin x - 4\sin^3 x$.
Therefore,$\frac{1}{4}(3\sin x - \sin 3x) = \sin^3 x$.
The expression becomes $\mathop {Limit}\limits_{x \to \frac{\pi }{2}} \frac{\sin x}{\cos^{-1}[\sin^3 x]}$.
As $x \to \frac{\pi}{2}$,$\sin x \to 1$,so $\sin^3 x \to 1$.
Since $\sin^3 x < 1$ for $x < \frac{\pi}{2}$,the greatest integer $[\sin^3 x] = 0$.
Thus,the limit is $\mathop {Limit}\limits_{x \to \frac{\pi}{2}^-} \frac{\sin x}{\cos^{-1}(0)} = \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi}$.
For $x > \frac{\pi}{2}$,$\sin x < 1$,so $\sin^3 x < 1$,and $[\sin^3 x] = 0$.
The limit is $\frac{2}{\pi}$ from both sides.

(A) Given the limit: $\mathop {Lim}\limits_{x \to 0} \frac{\ln(3 + x) - \ln(3 - x)}{x}$
Using the property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$,we get:
$\mathop {Lim}\limits_{x \to 0} \frac{\ln(\frac{3+x}{3-x})}{x}$
Using the standard limit $\mathop {Lim}\limits_{u \to 0} \frac{\ln(1+u)}{u} = 1$,we rewrite the expression:
$\mathop {Lim}\limits_{x \to 0} \frac{\ln(1 + \frac{x}{3}) - \ln(1 - \frac{x}{3})}{x}$
$= \mathop {Lim}\limits_{x \to 0} \left[ \frac{\ln(1 + \frac{x}{3})}{x} - \frac{\ln(1 - \frac{x}{3})}{x} \right]$
$= \mathop {Lim}\limits_{x \to 0} \left[ \frac{1}{3} \cdot \frac{\ln(1 + \frac{x}{3})}{\frac{x}{3}} - (-\frac{1}{3}) \cdot \frac{\ln(1 - \frac{x}{3})}{-\frac{x}{3}} \right]$
$= \frac{1}{3}(1) + \frac{1}{3}(1) = \frac{2}{3}$
Thus,$k = \frac{2}{3}$.

(C) To evaluate $f(x) = \lim_{n \to \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$:
Case $1$: If $|x| < 1$,then $x^{2n} \to 0$ as $n \to \infty$. Thus,$f(x) = \frac{0 - 1}{0 + 1} = -1$.
Case $2$: If $|x| > 1$,then divide numerator and denominator by $x^{2n}$,so $f(x) = \lim_{n \to \infty} \frac{1 - 1/x^{2n}}{1 + 1/x^{2n}} = \frac{1 - 0}{1 + 0} = 1$.
Case $3$: If $x = 1$,$f(1) = \frac{1 - 1}{1 + 1} = 0$. If $x = -1$,$f(-1) = \frac{1 - 1}{1 + 1} = 0$.
Thus,$f(x) = -1$ for $|x| < 1$,$0$ for $|x| = 1$,and $1$ for $|x| > 1$.
Now,consider $u(x) = \text{sgn}(|x| - 1)$:
If $|x| < 1$,$|x| - 1 < 0$,so $\text{sgn}(|x| - 1) = -1$.
If $|x| = 1$,$|x| - 1 = 0$,so $\text{sgn}(|x| - 1) = 0$.
If $|x| > 1$,$|x| - 1 > 0$,so $\text{sgn}(|x| - 1) = 1$.
Therefore,$f(x) = u(x)$.

(D) Let $x = \frac{\pi}{4} + h$. As $x \to \frac{\pi}{4}$,$h \to 0$.
Then $\cos x + \sin x = \cos(\frac{\pi}{4} + h) + \sin(\frac{\pi}{4} + h) = \sqrt{2} \cos h$.
Also,$1 - \sin 2x = 1 - \sin(\frac{\pi}{2} + 2h) = 1 - \cos 2h = 2 \sin^2 h$.
The expression becomes $\lim_{h \to 0} \frac{2\sqrt{2} - (\sqrt{2} \cos h)^3}{2 \sin^2 h} = \lim_{h \to 0} \frac{2\sqrt{2}(1 - \cos^3 h)}{2 \sin^2 h} = \sqrt{2} \lim_{h \to 0} \frac{(1 - \cos h)(1 + \cos h + \cos^2 h)}{\sin^2 h}$.
Using $\lim_{h \to 0} \frac{1 - \cos h}{\sin^2 h} = \lim_{h \to 0} \frac{1 - \cos h}{1 - \cos^2 h} = \lim_{h \to 0} \frac{1}{1 + \cos h} = \frac{1}{2}$.
Thus,the limit is $\sqrt{2} \times \frac{1}{2} \times (1 + 1 + 1) = \sqrt{2} \times \frac{3}{2} = \frac{3}{\sqrt{2}}$.

(B) Let $f(x) = \frac{\sin \{x - 10\}}{\{10 - x\}}$. Since $\{x - 10\} = \{x\}$ and $\{10 - x\} = 1 - \{x\}$ for $x \notin \mathbb{Z}$,the expression becomes $\frac{\sin \{x\}}{1 - \{x\}}$.
For $RHL$: $\mathop {\text{Lim}}\limits_{x \to 8^+} \{x\} = 0$,so $\mathop {\text{Lim}}\limits_{x \to 8^+} \frac{\sin \{x\}}{1 - \{x\}} = \frac{\sin(0)}{1 - 0} = 0$. Thus,$RHL$ exists.
For $LHL$: $\mathop {\text{Lim}}\limits_{x \to 8^-} \{x\} = 1$,so $\mathop {\text{Lim}}\limits_{x \to 8^-} \frac{\sin \{x\}}{1 - \{x\}} = \frac{\sin(1)}{0^+}$,which tends to $\infty$. Thus,$LHL$ does not exist.

(A) Let the numerator be $S_n = \sum_{k=1}^{n} k^2(n - k + 1)$.
$S_n = (n+1) \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k^3$.
We know that $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$.
The denominator is $D_n = \sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$.
Thus,the limit is $\lim_{n \to \infty} \frac{(n+1) \frac{n(n+1)(2n+1)}{6} - \frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)^2}{4}}$.
Dividing by $\frac{n^2(n+1)^2}{4}$:
$= \lim_{n \to \infty} \left( \frac{(n+1) n(n+1)(2n+1)}{6} \cdot \frac{4}{n^2(n+1)^2} - 1 \right)$.
$= \lim_{n \to \infty} \left( \frac{4n(n+1)^2(2n+1)}{6n^2(n+1)^2} - 1 \right)$.
$= \lim_{n \to \infty} \left( \frac{4(2n+1)}{6n} - 1 \right) = \frac{8}{6} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

(A) Let the given limit be $L = \mathop {\text{Limit}}\limits_{x \to \infty } \frac{{\cot ^{ - 1}}\left( \frac{\log _a x}{x^a} \right)}{{\sec ^{ - 1}}\left( \frac{a^x}{\log _a x} \right)}$.
As $x \to \infty$,$\frac{\log _a x}{x^a} \to 0$ because exponential growth dominates logarithmic growth for $a > 1$.
Thus,the numerator approaches $\cot ^{ - 1}(0) = \pi / 2$.
As $x \to \infty$,$\frac{a^x}{\log _a x} \to \infty$ because exponential growth dominates logarithmic growth.
Thus,the denominator approaches $\sec ^{ - 1}(\infty) = \pi / 2$.
Therefore,$L = \frac{\pi / 2}{\pi / 2} = 1$.

(C) Case $1$: If $x \in \mathbb{Z}$,then $\sin(\pi x) = 0$.
Thus,$f(x) = \lim_{n \to \infty} \frac{1}{1 + n(0)} = 1$.
Case $2$: If $x \notin \mathbb{Z}$,then $\sin^2(\pi x) > 0$.
As $n \to \infty$,$n \sin^2(\pi x) \to \infty$.
Thus,$f(x) = \lim_{n \to \infty} \frac{1}{1 + n \sin^2(\pi x)} = 0$.
Therefore,$f(x) = \begin{cases} 1, & x \in \mathbb{Z} \\ 0, & x \notin \mathbb{Z} \end{cases}$.
Since the function jumps between $0$ and $1$ at every integer value,$f$ is discontinuous for all $x \in \mathbb{Z}$.

(B) We need to find $\lim_{x \to 0} f(\phi(x))$.
First,let us evaluate $\lim_{x \to 0} \phi(x)$.
For $x \neq 0$ and $x \neq 3$,$\phi(x) = x^2 + 1$.
As $x \to 0$,$\phi(x) \to 0^2 + 1 = 1$.
Since $\phi(x)$ approaches $1$ as $x \to 0$ (and $1 \neq 0, 1 \neq 3$),we look at the definition of $f(x)$ near $1$.
Since $1$ is not of the form $n\pi$ for any integer $n \geq 0$,$f(x) = 3$ for values of $x$ near $1$.
Therefore,$\lim_{x \to 0} f(\phi(x)) = f(1) = 3$.

(D) As $x \to 0^-$,$\tan x$ approaches $0$ from the negative side,i.e.,$-1 < \tan x < 0$.
Therefore,$[\tan x] = -1$.
Substituting this into the limit:
$\mathop {Lim}\limits_{x \to {0^ - }} \sin^{-1}([\tan x]) = \sin^{-1}(-1) = -\frac{\pi}{2}$.
So,$l = -\frac{\pi}{2}$.
We need to find $\{l\} = \{-\frac{\pi}{2}\}$.
Using the property $\{x\} = x - [x]$,we have $\{ -\frac{\pi}{2} \} = -\frac{\pi}{2} - [-\frac{\pi}{2}]$.
Since $-1.57 < -\frac{\pi}{2} < -1$,the greatest integer $[-\frac{\pi}{2}] = -2$.
Thus,$\{ -\frac{\pi}{2} \} = -\frac{\pi}{2} - (-2) = 2 - \frac{\pi}{2}$.

(A) For $x \to \infty$,$f(x) = \frac{\ln(e^x(1 + x^2/e^x))}{\ln(e^{2x}(1 + x^4/e^{2x}))} = \frac{x + \ln(1 + x^2/e^x)}{2x + \ln(1 + x^4/e^{2x})}$.
Dividing numerator and denominator by $x$,we get $\lim_{x \to \infty} f(x) = \frac{1 + 0}{2 + 0} = \frac{1}{2}$,so $l = 1/2$.
For $x \to -\infty$,let $t = -x$,so $t \to \infty$. Then $f(-t) = \frac{\ln(t^2 + e^{-t})}{\ln(t^4 + e^{-2t})}$.
As $t \to \infty$,$e^{-t} \to 0$ and $e^{-2t} \to 0$.
Thus,$m = \lim_{t \to \infty} \frac{\ln(t^2)}{\ln(t^4)} = \lim_{t \to \infty} \frac{2 \ln(t)}{4 \ln(t)} = \frac{2}{4} = \frac{1}{2}$.
Since $l = 1/2$ and $m = 1/2$,we have $l = m$.

(C) We want to evaluate $L = \mathop {Lim}\limits_{n \to \infty } \cos \left( {\pi \sqrt {{n^2} + n} } \right)$.
Since $n$ is an integer,we can write $\cos \left( {\pi \sqrt {{n^2} + n} } \right) = \cos \left( {n\pi - n\pi + \pi \sqrt {{n^2} + n} } \right) = \cos \left( {n\pi - \pi \left( {n - \sqrt {{n^2} + n} } \right)} \right)$.
Using the identity $\cos(n\pi - \theta) = (-1)^n \cos(\theta)$,we have $L = \mathop {Lim}\limits_{n \to \infty } (-1)^n \cos \left( {\pi \left( {n - \sqrt {{n^2} + n} } \right)} \right)$.
Now,rationalize the expression inside the cosine:
$n - \sqrt {{n^2} + n} = \frac{{(n - \sqrt {{n^2} + n} )(n + \sqrt {{n^2} + n} )}}{{n + \sqrt {{n^2} + n} }} = \frac{{{n^2} - ({n^2} + n)}}{{n + \sqrt {{n^2} + n} }} = \frac{{ - n}}{{n + n\sqrt {1 + 1/n} }} = \frac{{ - 1}}{{1 + \sqrt {1 + 1/n} }}$.
As $n \to \infty$,this expression approaches $\frac{{ - 1}}{{1 + 1}} = -1/2$.
Thus,$L = \mathop {Lim}\limits_{n \to \infty } (-1)^n \cos \left( {\pi \left( -\frac{1}{2} \right)} \right) = \mathop {Lim}\limits_{n \to \infty } (-1)^n \cos \left( -\frac{\pi}{2} \right) = \mathop {Lim}\limits_{n \to \infty } (-1)^n \cdot 0 = 0$.

(C) To evaluate the limit $L = \lim \limits _{x \to 0} \frac{{{{(\sin x - \tan x)}^2} - {{(1 - \cos 2x)}^4} + {x^5}}}{{7\cdot{{({{\tan }^{ - 1}}x)}^7}\, + {{({{\sin }^{ - 1}}x)}^6}+ 3{{\sin }^5}x}}$,we analyze the leading terms as $x \to 0$.
Numerator:
$(\sin x - \tan x)^2 = (x - \frac{x^3}{6} - (x + \frac{x^3}{3}))^2 + O(x^8) = (-\frac{x^3}{2})^2 + O(x^8) = \frac{x^6}{4} + O(x^8)$.
$(1 - \cos 2x)^4 = (1 - (1 - \frac{(2x)^2}{2}))^4 = (2x^2)^4 = 16x^8$.
Thus,the numerator is $\frac{x^6}{4} - 16x^8 + x^5 \approx x^5$ as $x \to 0$.
Denominator:
$7(\tan^{-1} x)^7 + (\sin^{-1} x)^6 + 3(\sin x)^5 \approx 7x^7 + x^6 + 3x^5$.
The leading term is $3x^5$.
Limit $L = \lim \limits _{x \to 0} \frac{x^5}{3x^5} = \frac{1}{3}$.

(B) For $l = \mathop {\lim}\limits_{x \to 0} \frac{[x]^2}{x^2}$:
As $x \to 0^+$,$[x] = 0$,so $\frac{[x]^2}{x^2} = 0$. Thus,$\mathop {\lim}\limits_{x \to 0^+} \frac{[x]^2}{x^2} = 0$.
As $x \to 0^-$,$[x] = -1$,so $\frac{[x]^2}{x^2} = \frac{1}{x^2} \to \infty$. Thus,$l$ does not exist.
For $m = \mathop {\lim}\limits_{x \to 0} \frac{[x^2]}{x^2}$:
As $x \to 0$,$x^2$ is a small positive value approaching $0$ from the right $(0 < x^2 < 1)$.
Therefore,$[x^2] = 0$ for all $x \neq 0$ in the neighborhood of $0$.
Thus,$\mathop {\lim}\limits_{x \to 0} \frac{0}{x^2} = 0$. So,$m$ exists and $m = 0$.
Conclusion: $m$ exists but $l$ does not.

(D) Let $f(x) = \frac{\tan(\{x\} - 1) \sin\{x\}}{\{x\}(\{x\} - 1)}$.
For the right-hand limit $(x \to 0^+)$,$\{x\} = x$ where $x$ is a small positive value $h$:
$\mathop {\text{Limit}}\limits_{h \to 0^+} \frac{\tan(h - 1) \sin h}{h(h - 1)} = \frac{\tan(-1)}{-1} \times \mathop {\text{Limit}}\limits_{h \to 0^+} \frac{\sin h}{h} = \tan 1 \times 1 = \tan 1$.
For the left-hand limit $(x \to 0^-)$,$\{x\} = 1 + x$ where $x$ is a small negative value $-h$ $(h > 0)$:
$\mathop {\text{Limit}}\limits_{h \to 0^+} \frac{\tan(1 - h - 1) \sin(1 - h)}{(1 - h)(1 - h - 1)} = \mathop {\text{Limit}}\limits_{h \to 0^+} \frac{\tan(-h) \sin(1 - h)}{(1 - h)(-h)} = \frac{\sin 1}{1} = \sin 1$.
Since the right-hand limit $(\tan 1)$ is not equal to the left-hand limit $(\sin 1)$,the limit does not exist.

(D) We evaluate the limit: $\mathop {Lim}\limits_{x \to \infty } \frac{2 + 2x + \sin 2x}{(2x + \sin 2x)e^{\sin x}}$.
Divide the numerator and denominator by $x$:
$\mathop {Lim}\limits_{x \to \infty } \frac{\frac{2}{x} + 2 + \frac{\sin 2x}{x}}{(2 + \frac{\sin 2x}{x})e^{\sin x}}$.
As $x \to \infty$,$\frac{2}{x} \to 0$ and $\frac{\sin 2x}{x} \to 0$.
The expression simplifies to $\mathop {Lim}\limits_{x \to \infty } \frac{2}{2e^{\sin x}} = \frac{1}{e^{\sin x}}$.
Since $\sin x$ oscillates between $-1$ and $1$ as $x \to \infty$,the value of $e^{\sin x}$ oscillates between $e^{-1}$ and $e^1$.
Therefore,the limit does not exist.

(D) Option $A$: $f(x) = [[x]] - [2x - 1]$ at $c = 3$.
Right hand limit: $\lim_{h \to 0} f(3+h) = \lim_{h \to 0} [[3+h]] - [2(3+h)-1] = 3 - 5 = -2$.
Left hand limit: $\lim_{h \to 0} f(3-h) = \lim_{h \to 0} [[3-h]] - [2(3-h)-1] = 2 - 4 = -2$.
Since $LHL = RHL$,the limit exists.
Option $B$: $f(x) = [x] - x$ at $c = 1$.
Right hand limit: $\lim_{h \to 0} f(1+h) = \lim_{h \to 0} [1+h] - (1+h) = 1 - 1 = 0$.
Left hand limit: $\lim_{h \to 0} f(1-h) = \lim_{h \to 0} [1-h] - (1-h) = 0 - 1 = -1$.
Since $LHL \neq RHL$,the limit does not exist.
Option $C$: $f(x) = \{x\}^2 - \{-x\}^2$ at $c = 0$.
Right hand limit: $\lim_{h \to 0} f(0+h) = \lim_{h \to 0} (h)^2 - (1-h)^2 = 0 - 1 = -1$.
Left hand limit: $\lim_{h \to 0} f(0-h) = \lim_{h \to 0} (1-h)^2 - (h)^2 = 1 - 0 = 1$.
Since $LHL \neq RHL$,the limit does not exist.
Therefore,both $(B)$ and $(C)$ are correct.

(D) For $f(x) = \frac{x(2^x - 1)}{1 - \cos x} = \frac{x(2^x - 1)}{2 \sin^2(x/2)}$.
As $x \to 0$,$f(x) = \frac{x(2^x - 1)}{2 (x/2)^2} = \frac{x(2^x - 1)}{x^2/2} = 2 \frac{2^x - 1}{x}$.
Using the standard limit $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$,we get $\lim_{x \to 0} f(x) = 2 \ln 2 = \ln 4$.
Thus,option $(C)$ is correct.
For $g(x) = 2^x \sin \left( \frac{\ln 2}{2^x} \right)$,as $x \to \infty$,$2^x \to \infty$,so $\frac{\ln 2}{2^x} \to 0$.
Let $u = \frac{\ln 2}{2^x}$,then $g(x) = \frac{\ln 2}{u} \sin(u) = \ln 2 \cdot \frac{\sin u}{u}$.
As $x \to \infty$,$u \to 0$,so $\lim_{x \to \infty} g(x) = \ln 2 \cdot 1 = \ln 2$.
Thus,option $(B)$ is correct.
Therefore,$(D)$ is the correct option.