Question

Find the points of discontinuity, if any, of the following functions: 

\[f\left( x \right) = \begin{cases}\frac{e^x - 1}{\log_e (1 + 2x)}, & \text{ if }x \neq 0 \\ 7 , & \text{ if } x = 0\end{cases}\]

Answer 1

Given: 

\[f\left( x \right) = \begin{cases}\frac{e^x - 1}{\log_e (1 + 2x)}, & \text{ if }x \neq 0 \\ 7 , & \text{ if } x = 0\end{cases}\]

We have 

\[\lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \frac{e^x - 1}{\log_e \left( 1 + 2x \right)} = \lim_{x \to 0} \frac{\left( \frac{e^x - 1}{x} \right)}{\left( \frac{2 \log_e \left( 1 + 2x \right)}{2x} \right)} = \frac{1}{2} \times \frac{\lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)}{\lim_{x \to 0} \left( \frac{\log_e \left( 1 + 2x \right)}{2x} \right)} = \frac{1}{2}\]

It is given that 

\[f\left( 0 \right) = 7\]

⇒  \[\lim_{x \to 0} f\left( x \right) \neq f\left( 0 \right)\]

Hence, the given function is discontinuous at x = 0 and continuous elsewhere.