MCQ
The function $f:R - \left\{ 0 \right\} \to R,$ given by $f\left( x \right) = \frac{1}{x} - \frac{2}{{{e^{2x}} - 1}}$ can be made continuous at $x=0$ by defining $f\left( 0 \right)$ .
- A$2$
- B$-1$
- C$0$
- ✓$1$
$f(0)=\mathop {\lim }\limits_{x \to 0} f(x)=\mathop {\lim }\limits_{x \to 0} \left[\frac{1}{x}-\frac{2}{e^{2 x}-1}\right]$
$=\mathop {\lim }\limits_{x \to 0} \frac{e^{2 x}-1-2 x}{x\left(e^{2 x}-1\right)}$
$\mathop {\lim }\limits_{x \to 0}\frac{1+2 x+\frac{(2 x)^{2}}{21}+\ldots .-1-2 x}{x\left(1+2 x+\frac{(2 x)^{2}}{21}+\ldots .-1\right)}$
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{4 x^{2}}{2 !}+\frac{8 x^{3}}{3 !}+\ldots .}{x^{2}\left(2+\frac{4 x}{2 !}+\frac{8 x^{2}}{3 !}+\ldots .\right)}$
$f(0)=\frac{\frac{4}{2 !}}{2}=\frac{4}{2 \times 2}$
$f(0)=1$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $S \geq \frac{1}{ e }$ $(B)$ $S \geq 1-\frac{1}{ e }$
$(C)$ $S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$ $(D)$ $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$