Question

Extend the definition of the following by continuity
$f\left(x\right)=\frac{1-\cos 7(x-\mathrm{\pi })}{5(x-\mathrm{\pi }{)}^2}$ at the point x = π.

Answer 1

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

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

​
$$\begin{gathered} \underset{x\to \mathrm{\pi }}{\lim }f\left(x\right)=f\left(\mathrm{\pi }\right) \\ \Rightarrow \underset{x\to \mathrm{\pi }}{\lim }\frac{1-\cos 7\left(\mathrm{x}-\mathrm{\pi }\right)}{5{\left(\mathrm{x}-\mathrm{\pi }\right)}^2}=f\left(\mathrm{\pi }\right) \\ \Rightarrow \frac{2}{5}\underset{x\to \mathrm{\pi }}{\lim }\frac{{\sin }^2\left({\displaystyle \frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}}\right)}{{\left(\mathrm{x}-\mathrm{\pi }\right)}^2}=f\left(\mathrm{\pi }\right) \\ \Rightarrow \frac{2}{5}\times \frac{49}{4}\underset{x\to \mathrm{\pi }}{\lim }\frac{{\sin }^2\left({\displaystyle \frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}}\right)}{{\displaystyle \frac{49}{4}}{\left(\mathrm{x}-\mathrm{\pi }\right)}^2}=f\left(\mathrm{\pi }\right) \\ \Rightarrow \frac{2}{5}\times \frac{49}{4}\underset{x\to \mathrm{\pi }}{\lim }\frac{{\sin }^2\left({\displaystyle \frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}}\right)}{{\left({\displaystyle \frac{7}{2}}\left(\mathrm{x}-\mathrm{\pi }\right)\right)}^2}=f\left(\mathrm{\pi }\right) \\ \Rightarrow \frac{2}{5}\times \frac{49}{4}\underset{x\to \mathrm{\pi }}{\lim }{\left[\frac{\sin \left({\displaystyle \frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}}\right)}{\left({\displaystyle \frac{7}{2}}\left(\mathrm{x}-\mathrm{\pi }\right)\right)}\right]}^2=f\left(\mathrm{\pi }\right) \\ \Rightarrow \frac{2}{5}\times \frac{49}{4}\times 1=f\left(\mathrm{\pi }\right) \\ \Rightarrow \frac{1}{5}\times \frac{49}{2}\times 1=f\left(\mathrm{\pi }\right) \\ \Rightarrow \frac{49}{10}=f\left(\mathrm{\pi }\right) \end{gathered}$$


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