Embibe Experts Solutions for Chapter: Limits, Exercise 2: Exercise 2

Let $f:R\to [0,\infty )$ be such that $\underset{x\to 5}{\lim }f\left(x\right)$ exists and $\underset{x\to 5}{\lim }\frac{\left(f\right(x){)}^2-9}{\sqrt{|x-5|}}=0.$ Then $\underset{x\to 5}{\lim }f\left(x\right)$ equals

The limiting value of $\frac{{\tan }^{2}x-2\tan x-3}{{\tan }^{2}x-4\tan x+3}$ at ${\tan }^{-1}x=3$ is

$\underset{x\to 0}{\lim }\frac{x \sqrt[3]{z^2-{\left(z-x\right)}^2}}{{\left(\sqrt[3]{8xz-4x^2}+\sqrt[3]{8xz}\right)}^4}$ is equal to

$\underset{h\to 0}{\lim }\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right)$ equals to

$\underset{x\to 0}{\lim }\frac{\cos \left(\sin x\right)-\cos x}{x^4}$ is equal to

If $g\left(x\right)$  is a polynomial satisfying $g\left(x\right)g\left(y\right)=g\left(x\right)+g\left(y\right)+g\left(xy\right)-2$ for all real $x$ and $y$ and $g\left(2\right)=5$, then $\underset{x\to 3}{\lim }g\left(x\right)$ is

The value of $\underset{x\to 0}{\lim }\frac{1-\cos x+2\sin x-{\sin }^{3}x-x^2+3x^4}{{\tan }^{3}x-6{\sin }^{2}x+x-5x^3}$ is

A function $f$ is said to be a rational function, if $f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}$, where $g\left(x\right)$ and $h\left(x\right)$ are polynomials such that $h\left(x\right)\ne 0$, then