The function f:R0R given by f(x)=1x-2e^2x-1 can be made continuous at x=0 by defining f(0)…

Question

The function $f:\mathbb{R}\setminus\{0\}	o\mathbb{R}$ given by $f(x)=\dfrac1x-\dfrac{2}{e^{2x}-1}$ can be made continuous at $x=0$ by defining $f(0)$ as:

Accepted Answer

$$f(0)=\lim_{x\to0}\frac{(e^{2x}-1)-2x}{x(e^{2x}-1)}$$
Applying L'Hopital's rule twice: numerator's second derivative at $0$ over denominator's leads to $\displaystyle\lim_{x\to0}\frac{4e^{2x}}{4(xe^{2x}+e^{2x})}=\frac{4}{4}=1$.

Solutions · 1