Search any question & find its solution
Question:
Answered & Verified by Expert
The function $f:R \sim\{0\} \rightarrow R$ given by
$f(x)=\frac{1}{x}-\frac{2}{e^{2 x}-1}$
can be made continuous at $x=0$ by defining $f(0)$ as
Options:
$f(x)=\frac{1}{x}-\frac{2}{e^{2 x}-1}$
can be made continuous at $x=0$ by defining $f(0)$ as
Solution:
2144 Upvotes
Verified Answer
The correct answer is:
$1$
$1$
$\lim _{x \rightarrow 0} \frac{1}{x}-\frac{2}{e^{2 x}-1}$
$\lim _{x \rightarrow 0} \frac{e^{2 x}-1-2 x}{x\left(e^{2 x}-1\right)}$
$\lim _{x \rightarrow 0} \frac{2 e^{2 x}-2}{\left(e^{2 x}-1\right)+2 x e^{2 x}}$
$\lim _{x \rightarrow 0} \frac{4 e^{2 x}}{4 e^{2 x}+4 x e^{2 x}}=1$.
$\lim _{x \rightarrow 0} \frac{e^{2 x}-1-2 x}{x\left(e^{2 x}-1\right)}$
$\lim _{x \rightarrow 0} \frac{2 e^{2 x}-2}{\left(e^{2 x}-1\right)+2 x e^{2 x}}$
$\lim _{x \rightarrow 0} \frac{4 e^{2 x}}{4 e^{2 x}+4 x e^{2 x}}=1$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.