Embibe Experts Solutions for Chapter: Limits, Exercise 1: KCET (UG) 2018

The right hand and left hand limit of the function $\left\{\begin{array}{l}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}, \mathrm{if} x\ne 0\\ 0 , \mathrm{if} x=0\end{array} \right.$ are respectively

$\underset{x\to 0}{\lim }\left(\frac{\tan x}{\sqrt{2x+4}-2}\right)$ is equal to

$\sum _{r=1}^n(2r-1)=x$ then $\underset{n\to \infty }{\lim }\left[\frac{1^3}{x^2}+\frac{2^3}{x^2}+\frac{3^3}{x^2}+\dots +\frac{n^3}{x^2}\right]=$

The value of $\underset{x\to 0}{\lim }\frac{\left|x\right|}{x}$ is:

Consider the following statements:
Statement $1:$$\underset{x\to 1}{\lim }\left[\frac{a{x}^2+bx+c}{c{x}^2+bx+a}\right]$ is $1$ (where $a+b+c\ne 0$)

Statement $2:$ $\underset{x\to -2}{\lim }\left[\frac{{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{2}}}{x+2}\right]$ is $\frac{1}{4}$.