Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The value of  <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{e^{1 / x}-1}{e^{1 / x}+1}, x \neq 0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightar

The value of  \mathrm{\lim _{x \rightarrow 0} \frac{e^{1 / x}-1}{e^{1 / x}+1}, x \neq 0} is
 

Option: 1

0


Option: 2

-1


Option: 3

1


Option: 4

none of these.


Answers (1)

best_answer

Let \mathrm{f(x)=\frac{e^{1 / x}-1}{e^{1 / x}+1}}. Then

\mathrm{\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) =\lim _{h \rightarrow 0} \frac{e^{1 / h-1}}{e^{1 / h}+1}=\lim _{h \rightarrow 0} \frac{1-e^{-1 / h}}{1+e^{-1 / h}} }

\mathrm{ =1}

and,  \mathrm{\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} \frac{e^{-1 / h-1}}{e^{-1 / h+1}}=-1}

Hence, \mathrm{\lim _{x \rightarrow 0} f(x)} does not exist.

Hence option 4 is correct.

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech Admissions 2025 LIVE: JEE Main, MET registration underway; how to apply
anam-khan

JEE Main 2025 Registration: NTA issues advisory for Aadhaar authentication, name mismatch issues
anam-khan

BTech Admissions 2025: JEE Main registration last date, direct link

Latest Question