Let f(x)=e^2x + x²
Since exponential function e^2x and algebraic function (x²) is continuous for all \(x \in R\)
f(x)=e^2x + x² is continuous for all \(x \in R\).

By Sandwitch theorem \(lim_{x\rightarrow 0}x^2sin{1\over x}=0\) and f(0) = 0 by the definition of f(x). Hence it is continuous at x = 0. For other values it is clearly continuous and hence continuous in R.

\(lim_{x\rightarrow \infty}(1+{k\over x})^{m\over x} \)
put \({1\over x}=t\)
\(\therefore lim_{x\rightarrow \infty}(1+{k\over x})^{m\over x} =lim_{{1\over x}\rightarrow\infty}(1+kt)^{mt}\)\(=lim_{t\rightarrow 0}(1+0)^{m(0)}=lim_{t\rightarrow 0}(1)^{0}=1\)

Given \(f(x)={x^2-4\over (x^2+4x+4)(x+3)}at \ x=-2\)

\(=lim_{x\rightarrow -2^-}{x-2\over (x+2)(x+3)}\)
\(={Negative \over 0(Negative)}=\infty\)
\(\therefore f(-2^-)\rightarrow \infty as \ x\rightarrow 2^-\)
\(f(-2)^+=lim_{x\rightarrow -2^+}{(x+2)(x-2)\over (x+2)(x+3)}\)
\(=lim_{x\rightarrow -2^+}{x-2\over (x+2)(x+3)}\)
\(={Negative \over 0(Positive)}\rightarrow-\infty\)
\(\therefore f(-2^+)\rightarrow -\infty as \ x\rightarrow 2^+\).

\(\frac{1}{x-3}\) can be made arbitrarily large by choosing x suficiently close to 1 on the right side but not equal to 1.
\(\therefore \frac{1}{x-3}\) does not approach any value when x approaches 3 from the right.
\(\therefore \lim _{x \rightarrow 3^{+}} \frac{1}{x-3}=x\) and hence the limit does not exist.

At x = 1, the value of the curve on the y-axis is 3.
\(\therefore lim_{x\rightarrow1}f(x)=3\)

At x = 2, the value of the curve on y-axis is 2.
\(\therefore lim_{x\rightarrow2}f(x)=2\)

\(lim_{x\rightarrow1}(x^2+2)\)
Atx = 1, the value of the curve on y-axis is 3.
\(\therefore lim_{x\rightarrow1}(x^2+2)=3\)

\(lim_{x\rightarrow3}(4-x)\)

At x = 3, the value of the curve on y-axis is 1.
\(\therefore lim_{x\rightarrow3}(4-x)=1\)

Let \( f(x)=\frac{\cos x-1}{x} \)
\(\therefore \lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0\)

Let \( f(x)=\frac{\sqrt{1-x}-2}{x+3} \)
\(\therefore \lim _{x \rightarrow-3} \frac{\sqrt{1-x}-2}{x+3}=-0.250 \)

Let \( f(x)=\frac{x-2}{x^2-x-2}=\frac{x-2}{(x-2)(x+1)}=\frac{1}{x+1} \)
\( \therefore \lim _{x \rightarrow 2} \frac{x-2}{\dot{x}^2-x-2}=\lim _{x \rightarrow 2} \frac{1}{x+1}=\frac{1}{3}=0 . \overline{3} \)

No. f(x) = \(\sqrt{x}\) is not even defined for x < 0.

Therefore as x \(\rightarrow 0^-,lim_{x\rightarrow0^-}\sqrt{x}\) does not exist.
However, \(lim_{x\rightarrow0^+}\sqrt{x}=0.\) Therefore \(lim_{x\rightarrow0}\sqrt{x}\) does not exist.

For minimum light it is enough to find the limit of the function when x\(\rightarrow\)0⁺.
\(lim_{x\rightarrow 0^+}f(x)=lim_{x\rightarrow 0^+}{160x^{-0.4}+90\over 4x^{-0.4}+15}=lim_{x\rightarrow0^+}{160+90x^{0.4}\over 4+15x^{0.4}}\)
\(={160\over 4}=40mm\).

\(f(x)= \begin{cases}\sin x, & x<0 \\ 1-\cos x, & 0 \leq x \leq \pi \\ \cos x, & x>\pi\end{cases}\)
To sketch the graph of y = f(x)
(i) Draw y = sin x where x < 0
(ii) To draw y = \(1-\cos x, 0 \leq x \leq \pi\)
Draw \(y=\cos x, 0 \leq x \leq \pi\)and reflect it on x-axis and then lift up by 1 unit.
(OR)
Putting x = 0, y = 1 - cos 0 = 1 -1 = 0
Putting \(x=\pi / 2, y=1-\cos \pi / 2=1-0=1\)
Putting \(x=\pi, y=1-\cos \pi=1+1=2\)
\(\therefore y=1-\cos x\) passes through \((0,0),\left(\frac{\pi}{2}, 1\right)\), \((\pi, 2)\)
(iii) Draw y = cos X where x >\(\pi\)
when \(x=\pi, y=\cos \pi=-1\)

\( \lim _{x \rightarrow \pi^{-}} f(x)=2 \)
\( \lim _{x \rightarrow \pi^{+}} f(x)=-1 \)
\( \therefore \lim _{x \rightarrow \pi} f(x) \) dose not exist. 
\(\therefore \lim _{x \rightarrow x_0} f(x) \text { exists for } x \in R-\{\pi\} \)

​\(f(x)=\begin{cases} { x^ 2 } ,\quad x \le 2 \\ 8-{ 2x } ,\quad 2 < x < 4 \\ { 4 } ,\quad x\ge 4 \end{cases}\)​
To sketch the graph of y = f(x)
(i) y = x² is an upward parabola with vertex at origin 
when x = 2, y = (2)² = 4
Draw the graph y = x² where \(x \leq 2\)
(ii) y = 8 - 2x is a straight line.
when x = 2, y = 8 - 4 = 4
when x = 4, y = 8 - 8 = 0
Join (2, 4) and (4, 0) by a line segment and delete the points (2, 4) and (4, 0)
(iii) y = 4 is a horizontal line with y intercept 4.
when x = 4, y = 4
Draw y = 4 where \(k \geq 4\)

\( \lim _{x \rightarrow 4^{-}} f(x)=0 \ and\ \lim _{x \rightarrow \mathbb{4}^{+}} f(x)=4 \)
\( \therefore \lim _{x \rightarrow 4} f(x) \) does not exist.
\( \therefore \lim _{x \rightarrow x_0} f(x) exists\ for\ x_0 \in R-\{4\} \)

Let f(x) = {\(\begin{matrix} { x }^{ 3 }-3, & if\quad x\le 2 \\ { x }^{ 2 }+1, & if\quad x>2 \end{matrix}\)
\(lim_{x \rightarrow 2^-}f(x)=lim_{x\rightarrow2^-}(x^3-3)=8-3=5\)
\(lim_{x \rightarrow 2^+}f(x)=lim_{x\rightarrow2^+}x^2+1=4+1=5\)
Also,f(2) = x³-3 = 2³-3 = 8-3 = 5
\(\therefore f(x)\) is continuous in R.

One can tabulate values of x near 0 (from either side) and conclude \(f(x)={1\over x^2+x^3}\) grows without bound and hence \(f(x)\rightarrow \infty as \ x\rightarrow 0.\)
To calculate this limit without making a table, we first divide the numerator and denominator by x². This division can be done, since in the calculation of the limit x ≠ 0 and hence x² ≠ 0. We can have
\(lim_{x\rightarrow 0}{1\over x^2+x^3}=lim_{x\rightarrow 0}{{1\over x^2}\over{x^2+x^3\over x^2}}={lim_{x\rightarrow}({1\over x^2})\over lim_{x\rightarrow0}(1+x)}\)
Now \({1\over 8}\rightarrow \infty as \ x\rightarrow 0\) and \(lim_{x\rightarrow0}(1+x)=1\) .
Thus the numerator grows without bound while the denominator approaches 1, implying that \({1\over (x^2+x^3)}\) does tend to infinity.

\(f(x) =\frac{x-|x|}{x}, x \neq 0 \)
\(=1-\frac{|x|}{x} \)
\(= \begin{cases}1-\frac{(-x)}{x}, & x<0 \\ 1-\frac{x}{x}, & x>0\end{cases} \)
\(= \begin{cases}1+1, & x<0 \\ 1-1, & x>0\end{cases} \)
\(= \begin{cases}2, & x<0 \\ 0, & x>0\end{cases} \)
\(\text { Given: } f(0)=2 \text {. }\)
\(\therefore f(x)= \begin{cases}2, & x \leq 0 \\ 0, & x>0\end{cases}\)
\(\text { Obviously } f\left(0^{-}\right)=2 \text { and } f\left(0^{+}\right)=0\)
\(\therefore f(x) \text { is not continuous at } x=0 \text {. }\)
\(\text { It is continuous elsewhere }\)

\(f(x)=\frac{|2 x-3|}{2 x-3} \text { is not defined at } x= \frac{3}{2}\)
\(\therefore f(x) \text { is not continuous at } x=\frac{3}{2}\)

\(\text { WKT }\left\lfloor k^{-}\right\rfloor=k-1 \text { and }\left\lfloor k^{+}\right\rfloor=k \text { where } k \text { is an integer }\)
\(\therefore \lim _{x \rightarrow k^{-}} x-\lfloor x\rfloor=k-(k-1)=1\)

\(\lim _{x \rightarrow 0^{-}} \frac{\sin x}{\sqrt{x^{2}}} =\lim _{x \rightarrow 0^{-}} \frac{\sin x}{|x|} \)
\(=\lim _{x \rightarrow 0^{-}} \frac{\sin x}{-x}(\because x<0) \)
= -1

\(\lim _{x \rightarrow 0} \frac{e^{\tan x}-e^{x}}{\tan x-x}=\lim _{x \rightarrow 0} e^{x}\left(\frac{e^{\tan x-x}-1}{\tan x-x}\right)\)
\(=e^{0}(1) \quad(\because \tan x-x \rightarrow 0)\)
=1

\(\lim _{n \rightarrow \infty}[\left.\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\cdots+\frac{n}{n^{2}}\right] \)
\(=\lim _{n \rightarrow \infty}\left[\frac{(1+2+3+\cdots+\dot{n})}{n^{2}}\right] \)
\(=\lim _{n \rightarrow \infty}\left[\frac{n(n+1)}{2 n^{2}}\right] \)
\(=\frac{1}{2} \lim _{n \rightarrow \infty}\left(\frac{n+1}{n}\right) \)
\(=\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=1 / 2(1+0)=1 / 2 \)

\(f(x)=x(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\)
\(=x(\pm 1) \left(\because\left[\frac{1}{x}\right\rfloor \text { is an integer }\right) \)
\(=\pm x \)
\(\therefore \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(\pm x)=0\)

\(\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}-1}}{2 x+1}=\lim _{x \rightarrow \infty} \frac{\sqrt{1-\frac{1}{x^{2}}}}{2+\frac{1}{x}}=\frac{\sqrt{1-0}}{2+0}=\frac{1}{2}\)

\(\lim _{x \rightarrow \pi / 2} \frac{2 x-\pi}{\cos x} =\lim _{x \rightarrow \pi / 2} \frac{2 x-\pi}{\sin \left(\frac{\pi}{2}-x\right)} \)
\(=\lim _{\left(\frac{\pi}{2}-x\right) \rightarrow 0} \frac{-2\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)}=-2 \)