The value of lim_{x rightarrow 0} frac{|x|}{x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We are given the limit $\lim_{x \rightarrow 0} \frac{|x|}{x}$.To evaluate this,we check the left-hand limit and the right-hand limit.For the right

Vedclass · Accepted Answer

Does not exist

Vedclass · Answer

(D) We are given the limit $\lim_{x ightarrow 0} \frac{|x|}{x}$.To evaluate this,we check the left-hand limit and the right-hand limit.For the right-hand limit $(x ightarrow 0^{+})$,$|x| = x$,so $\lim_{x ightarrow 0^{+}} \frac{x}{x} = 1$.For the left-hand limit $(x ightarrow 0^{-})$,$|x| = -x$,so $\lim_{x ightarrow 0^{-}} \frac{-x}{x} = -1$.Since the left-hand limit $
eq$ the right-hand limit,the limit does not exist.

The value of $\lim_{x \rightarrow 0} \frac{|x|}{x}$ is

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} - {b^x}}}{x}} \right) = $

$\lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$ is equal to

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \cdot \frac{1-\sin x}{(\pi-2 x)^3} = $

If $\mathop {Lim}\limits_{x \to 0} \frac{\ln(3 + x) - \ln(3 - x)}{x} = k$,the value of $k$ is

$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + \tan x}}{{1 + \sin x}}} \right)^{\text{cosec } x}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module