Question

Find the value f(0) so that the function $f\left(x\right)=\frac{1}{x}-\frac{2}{e^{2x}-1},x\ne 0$ is continuous at $x=0$ & examine the differentiability of f(x) at $x=0$

• A. $f\left(0\right)=0$, differentiable at $x=0$
• B. $f\left(0\right)=0$, not differentiable at $x=0$
• C. $f\left(0\right)=1$, differentiable at $x=0$
• D. $f\left(0\right)=1$,not differentiable at $x=0$

Answer 1

The correct option is C $f\left(0\right)=1$, differentiable at $x=0$

$f\left(x\right)=\frac{1}{x}-\frac{2}{e^{2x}-1}=\underset{x\to 0}{lim}\frac{e^{2x}-1-2x}{x(e^{2x}-1)}$
$=\underset{x\to 0}{lim}\frac{2e^{2x}-2}{e^{2x}-1+x\left(2e^{2x}\right)}$ (by L-hopital rule)
$=\underset{x\to 0}{lim}\frac{4e^{2x}}{2e^{2x}+2e^{2x}+4e^{2x}.x}=1\Rightarrow f\left(0\right)=1$
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{\frac{1}{0+h}-\frac{2}{2^{2h}-1}-1}{h}$
$=\underset{h\to 0}{lim}\frac{e^{2h}-1-2h-h(e^{2h}-1)}{h^2(e^{2h}-1)}$
$=\underset{h\to 0}{lim}\frac{1+2h+\frac{4h^2}{2!}.....-1-2h-h(e^{2h}-1)}{h^2(e^{2h}-1)}$
$=\underset{h\to 0}{lim}\frac{(2h^2+\frac{8h^3}{3!}+....)-h(2h+\frac{(2h{)}^2}{2!}+\frac{8h^3}{3!}....)}{h^2(2h+\frac{(2h{)}^2}{2!}+.....)}=-\frac{1}{3}$
Similarly $f^{\prime }\left(0^{-}\right)=-\frac{1}{3}$.