#### Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 10 Subquestion (iv)

Discontinuous

Hint:

For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Solution:

Given,

$f(x)=\left(\begin{array}{c} \frac{e^{x}-1}{\log (1+2 x)}, \text { if } x \neq 0 \\\\ 7, \text { if } x=0 \end{array}\right)$

We observe,

\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{e^{x}-1}{\log (1+2 x)} \\\\ &=\lim _{x \rightarrow 0} \frac{e^{x}-1}{\frac{2 x \log (1+2 x)}{2 x}} \end{aligned}        [Multiplying and dividing the denominator by  $2x$]

\begin{aligned} &=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\left(\frac{e^{x}-1}{x}\right)}{\left(\frac{\log (1+2 x)}{2 x}\right)} \\\\ \end{aligned}

$=\frac{1}{2} \frac{\left(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\right)}{\left(\lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x}\right)}=\frac{1}{2} \times \frac{1}{1}=\frac{1}{2}$

And  $f\left ( 0 \right )=7$

$\lim _{x \rightarrow 0} f(x) \neq f(0)$

Thus,  $f\left ( x \right )$ is discontinuous at $x=0$.