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

f(x) is continuous on [2, 5] and f(x) takes only rational talues means f(x) is a constant rational value for all values of \(x \in[2,5]\)
\(\text { Since } f(3)=12, f(4.5)=12\)

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

\(f(x)=\left\{\begin{array}{cc} x, & x \text { is irrational } \\ 1-x, & x \text { is rational } \end{array}\right.\)
\(\text { Let } C \in R \text { be any element. }\)
\(\text { As } x \text { is getting close to } C, f(x) \text { oscillates between }C \text { and } 1 \text { - } C \text {. }\)
\(\therefore f(x) \text { is discontinuous everywhere }\)

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

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

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

\(\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x}=\lim _{x \rightarrow 0} \frac{e^{x}-1}{x} \quad(\text { replacing } \sin x \text { by } x)\)
= 1

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

We know that,
\(\sin \alpha-\cos \alpha =\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin \alpha-\frac{1}{\sqrt{2}} \cos \alpha\right] \)
\(=\sqrt{2}\left[\cos \frac{\pi}{4} \sin \alpha-\sin \frac{\pi}{4} \cos \alpha\right] \)
\(=\sqrt{2} \sin \left(\alpha-\frac{\pi}{4}\right) \)
\(\therefore \lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}} =\lim _{\alpha-\frac{\pi}{4} \rightarrow 0} \frac{\sqrt{2} \sin \left(\alpha-\frac{\pi}{4}\right)}{\left(\alpha-\frac{\pi}{4}\right)}=\sqrt{2} \)