The right hand and left hand limit of the fun | vedclass

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

Question

(B) Given $f(x) = \begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & x 
eq 0 \ 0, & x=0 \end{cases}$ Right hand limit $(RHL)$: $\lim_{x ightarrow 0^

Vedclass · Accepted Answer

$1$ and $-1$

Vedclass · Answer

(B) Given $f(x) = \begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & x 
eq 0 \ 0, & x=0 \end{cases}$ Right hand limit $(RHL)$: $\lim_{x ightarrow 0^{+}} f(x) = \lim_{h ightarrow 0} f(0+h) = \lim_{h ightarrow 0} \frac{e^{1/h}-1}{e^{1/h}+1}$ Dividing numerator and denominator by $e^{1/h}$: $\lim_{h ightarrow 0} \frac{1 - e^{-1/h}}{1 + e^{-1/h}} = \frac{1 - 0}{1 + 0} = 1$ Left hand limit $(LHL)$: $\lim_{x ightarrow 0^{-}} f(x) = \lim_{h ightarrow 0} f(0-h) = \lim_{h ightarrow 0} \frac{e^{-1/h}-1}{e^{-1/h}+1}$ As $h ightarrow 0^{+}$,$e^{-1/h} ightarrow 0$: $\frac{0 - 1}{0 + 1} = -1$ Thus,the $RHL$ is $1$ and the $LHL$ is $-1$.

The right hand and left hand limit of the function $f(x)$ are respectively:
$f(x)=\begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{x\sqrt {{y^2} - {{(y - x)}^2}} }}{{{{(\sqrt {8xy - 4{x^2}} + \sqrt {8xy} )}^3}}}$ is equal to

If $[x]$ denotes the greatest integer $\leq x$,then $\lim_{n \rightarrow \infty} \frac{1}{n^3} \{[1^2 x] + [2^2 x] + [3^2 x] + \ldots + [n^2 x] \} = $

Find the limit: $\mathop {\lim }\limits_{x \to 1} \frac{x^{2}+1}{x+100}$

$[x]$ denotes the greatest integer less than or equal to $x$. If $\{x\}=x-[x]$ and $\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x+[x])}{2-\{x\}}=\theta$,then $\sin \theta+\cos \theta=$

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} (\text{cosec}\,x - \cot x)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The right hand and left hand limit of the function $f(x)$ are respectively: $f(x)=\begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$