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

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

Answers (1)

Answer:

                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.

 

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