Let f(x) = cot x + tan x
cot x is not continuous in multiples of \(\pi\) and tan x is not continuous in (2n+1)\({\pi\over 2}\).
\(\therefore \) f(x) = cot x + tan x is not continuous in (2n+1)\({\pi\over 2}\)+\({\pi\over 2}\),
\(\Rightarrow\) f(x) is continuous in R \(-{n\pi \over2}, n \in Z.\)

Let f(x)=x + sin x
Since the algebraic function x is continuous for all \(x \in R\) and the circular function sin x is continuous for all \(x \in R\).
f(x)=x + sin x is continuous for all \(x \in R\).

\(lim_{x\rightarrow \infty}(1+{k\over x})^{m\over x} \)
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 \(f(x)=tan \ x \ at \ x={\pi\over 2}.\)
lim f(x) = lim tan x = \(\infty\)
\( x\rightarrow{\pi^-\over 2}\)   \( x\rightarrow{\pi^-\over 2}\)
\(\therefore lim_{ x\rightarrow{\pi^-\over 2}} f(x)\rightarrow \infty \ as \ x\rightarrow{\pi^-\over 2}\)
lim f(x) = lim tan x
\( x\rightarrow{\pi^+\over 2}\)   \( x\rightarrow{\pi^+\over 2}\)
= \(-\infty\)     [\(\because \ tan \pi =tan (90+90)=-cot 90=-\infty\)]
\(\therefore lim_{x\rightarrow{\pi^+\over 2}} f(x)\rightarrow -\infty \ as \ x\rightarrow{\pi^+\over 2}\)

\(lim_{x\rightarrow5}{\sqrt{x-1}-2\over x-5}\)
Multiplying and dividing by\((\sqrt{x-1}+2)\) we get,
\(lim_{x\rightarrow5}{\sqrt{x-1}-2\over x-5}\times {\sqrt{x-1}+2\over \sqrt{x-1}+2}\)\(=lim_{x\rightarrow5}{(x-1)-4\over x-5[\sqrt{x-1}+2]}\)

\(=lim_{x\rightarrow5}{1\over\sqrt{x-1}+2}={1\over\sqrt{5-1}+2}\)
\({1\over \sqrt{4}+2}={1\over2+2}={1\over 4}\)
\(lim_{x\rightarrow5}{\sqrt{x-1}-2\over x-5}={1\over4}\)

\(lim_{x\rightarrow1}{x^3-1\over x-1}=lim_{x\rightarrow1}{x^3-1\over x-1}=3(1)^{3-1}=3.\)

f(x) can be made arbitrarily close to 25 by choosing x sufficiently close to 8
but not equal to 8
Left limit of f(x) = Right limit of f(x) when x \(\rightarrow\) 8
\(lim_{x\rightarrow8^-}f(x)=lim_{x\rightarrow8^+}f(x)=25.\)
\(\therefore f(8^-)=f(8^+)=25\)

\(\frac{|x-5|}{x-5}\) approaches 1 on the right side of 5 and -1 on the left side of 5.
Hence, the limit does not exist.

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

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.

\(|x|= \begin{cases}-x & \text { if } x<0 \\ 0 & \text { if } x=0 \\ x & \text { if } x>0\end{cases}\)
If x > 0,then |x| =  x, which tends to 0 as
\(x \rightarrow 0\) from the right of 0. That is, \(\lim _{ x\rightarrow0^+}{|x| } =0\)
If x < 0, then |x| = - x which again tends to 0 as x\(\rightarrow\)0. from the left of 0. That is, \(\lim _{ x\rightarrow0^-}{|x| } =0\).
Thus, \(\lim _{ x\rightarrow0^-}{|x| } =0=\lim _{ x\rightarrow0^+}{|x| }.\)
Hence \(\lim _{ x\rightarrow0}{|x| } =0\).

Let x denote the number of kilograms bought per day and C denote the cost. Then,
\(C(x)= \begin{cases}0.16 x, & \text { if } 0 \leq x<100 \\ 0.14 x, & \text { if } x \geq 100\end{cases}\)
The sketch of this function
It is discontinuous at x = 100 since \(lim_{x\rightarrow100^-}c(x)=16\) and \(lim_{x\rightarrow100^+}c(x)=14\)
Note that C(100) = 14. Thus,\(lim_{x\rightarrow100^-}c(x)=16\neq 14=lim_{x\rightarrow100^+}C(x)=C(100).\)
Note also that the function jumps from one finite value 14 to another finite value 16.

(i) f(-5^-)
For x < - 5, |x + 5| = - (x + 5)
Thus  f(-5^-) = \(lim_{x\rightarrow-5^- {-(x+5)\over (x+5)}}=-1\)
(ii) f(-5⁺)
For x > - 5, |x + 5| = (x + 5)
Thus  f(-5⁺) = \(lim_{x\rightarrow-5^+{(x+5)\over (x+5)}}=1\)
Note that f( -5^-)  ≠ f( -5⁺). Hence the limit does not exist.

\(f(-1) =\lim _{x \rightarrow-1} f(x) \)
\(=\lim _{x \rightarrow-1} \frac{x^{2}-1}{x^{3}+1} \)
\(=\lim _{x \rightarrow-1} \frac{x^{2}-(-1)^{2}}{x^{3}-(-1)^{3}}=\frac{2(-1)^{2-1}}{3(-1)^{3-1}}=\frac{-2}{3} \)

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

\(\lim _{x \rightarrow 0} \frac{\sin p x}{\tan 3 x}=4 \Rightarrow \frac{p}{3}=4 \Rightarrow p=12\)

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