Find the points of discontinuity of $ f(x) =\frac{e^x-1}{\log_{e}(1+2x)}$ if $x\ne0$ and $f(0)=7$ Find the points of discontinuity:$$
f(x) = 
\begin{cases}
\frac{e^x-1}{\log_{e}(1+2x)} &amp; \text{if} \ x\ne0 \\ \\
7 &amp;\text{if} \ x=0 \ 
\end{cases}.
$$


I tried to solve it like this: $7$ being a constant functon is continuous and $\frac{e^x-1}{\log_{e}(1+2x)}$ is continuous.(Why???)

At $x=0$, $$\lim_{h\to 0}\frac{e^x-1}{\log_{e}(1+2x)}=???$$
and
$$\lim_{h\to 0}\frac{\log_{e}(1+2x)}{2x}=1 , \lim_{h\to 0}\frac{e^x-1}{x}=\log_{e}e=1  $$
I am not sure the second one is correct and I couldn't simplify $\lim_{h\to 0}\frac{e^x-1}{\log_{e}(1+2x)}$
$$\lim_{h\to 0}\frac{e^h-1}{\log_e(1+2)}= \lim_{h\to 0}\frac{\frac{e^h-1}{h}}{\frac{\log_e(1+2h)}{2h}}.\lim_{h\to 0}\frac{x}{2x}=\frac{1}{2}$$
and $$\lim_{h\to 0}\frac{\log_{e}(1+2x)}{2x}=1 , \lim_{h\to 0}\frac{e^x-1}{x}=\log_{e}e=1  $$
Therefore $0$ is the only point of discontinuity.
Originally posted by lab bhattacharjee above
$\log (1 + 2x)$ is not defined in the interval $(-∞,-1/2]$
hence, the function is discontinuous at every points in the interval $(-∞,-1/2] $ also.