Question

Discuss the continuity of the following functions at the indicated point(s): (iv) \[f\left( x \right) = \left\{ \begin{array}{l}\frac{e^x - 1}{\log(1 + 2x)}, if & x \neq a \\ 7 , if & x = 0\end{array}at x = 0 \right.\]

Answer 1

 Given:

\[f\left( x \right) = \binom{\frac{e^x - 1}{\log\left( 1 + 2x \right)}, if x \neq 0}{7, if x = 0}\]

We observe

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

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

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

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

And, 

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

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

Hence, f(x) is discontinuous at x = 0.