Question

The function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{1}{4^x-1},& x\ne 0\end{array}\\ \begin{array}{cc}0,& x=0\end{array}\end{array}$ is continuous

• A. everywhere except at $x=0$ and $x=\infty$
• B. nowhere
• C. everywhere
• D. everywhere expect at $x=0$

Answer 1

The correct option is D everywhere expect at $x=0$

To examine $f\left(x\right)$ at $x=0$ $($ as at rest of the point, it is continuous $)$.

$f\left(0\right)=0$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{1}{4^x-1}=\underset{h\to 0}{lim}\frac{1}{4^{0+h}-1}=\underset{h\to 0}{lim}\frac{1}{4^h-1}=\infty$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{1}{4^x-1}=\underset{h\to 0}{lim}\frac{1}{4^{0-h}-1}=\underset{h\to 0}{lim}\frac{1}{4^{-h}-1}=-\infty$