# Extend the definition of the following by continuity

Question:

Extend the definition of the following by continuity

$f(x)=\frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}$ at the point $x=\pi$

Solution:

Given:

$f(x)=\frac{1-\cos 7(\mathrm{x}-\pi)}{5(\mathrm{x}-\pi)^{2}}, \quad x=\pi$

If $f(x)$ is continuous at $x=\pi$, then

$\lim _{x \rightarrow \pi} f(x)=f(\pi)$

$\Rightarrow \lim _{x \rightarrow \pi} \frac{1-\cos 7(\mathrm{x}-\pi)}{5(\mathrm{x}-\pi)^{2}}=f(\pi)$

$\Rightarrow \frac{2}{5} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(\mathrm{x}-\pi)}{2}\right)}{(\mathrm{x}-\pi)^{2}}=f(\pi)$

$\Rightarrow \frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(\mathrm{x}-\pi)}{2}\right)}{\frac{49}{4}(\mathrm{x}-\pi)^{2}}=f(\pi)$

$\Rightarrow \frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(\mathrm{x}-\pi)}{2}\right)}{\left(\frac{7}{2}(\mathrm{x}-\pi)\right)^{2}}=f(\pi)$

$\Rightarrow \frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi}\left[\frac{\sin \left(\frac{7(x-\pi)}{2}\right)}{\left(\frac{7}{2}(x-\pi)\right)}\right]^{2}=f(\pi)$

$\Rightarrow \frac{2}{5} \times \frac{49}{4} \times 1=f(\pi)$

$\Rightarrow \frac{1}{5} \times \frac{49}{2} \times 1=f(\pi)$

$\Rightarrow \frac{49}{10}=f(\pi)$

Hence, the given function will be continuous at $x=\pi$, if $f(\pi)=\frac{49}{10}$.