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#### Please Solve RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 7 Maths Textbook Solution.

$x=0$ (Discontinuous)

Hint:

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

Solution:

Given,

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

Consider,

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)$

$=\lim _{x \rightarrow 0}\left(\frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}\right)$                                                                                      $\left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]$

\begin{aligned} &=\lim _{x \rightarrow 0}\left(\frac{2 \sin \frac{x}{2}}{4\left(\frac{x^{2}}{4}\right)}\right) \\\\ &=\lim _{x \rightarrow 0}\left(\frac{2\left(\sin \frac{x}{2}\right)^{2}}{4\left(\frac{x}{2}\right)^{2}}\right) \end{aligned}

$=\frac{2}{4} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}$                                                                                                                $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$\lim _{x \rightarrow 0} f(x)=\frac{1}{2} \times 1^{2}=\frac{1}{2}$

Given

$f\left ( 0 \right )=1$

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

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