Question

Extend the definition of the following by continuity 

\[f\left( x \right) = \frac{1 - \cos7 (x - \pi)}{5 (x - \pi )^2}\]  at the point x = π.

Answer 1

Given: 

\[f\left( x \right) = \frac{1 - \cos7\left( x - \pi \right)}{5 \left( x - \pi \right)^2}, x = \pi\]

If f(x) is continuous at x =

\[\pi\]then

\[\lim_{x \to \pi} f\left( x \right) = f\left( \pi \right)\]
\[ \Rightarrow \lim_{x \to \pi} \frac{1 - \cos7\left( x - \pi \right)}{5 \left( x - \pi \right)^2} = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \lim_{x \to \pi} \frac{\sin^2 \left( \frac{7\left( x - \pi \right)}{2} \right)}{\left( x - \pi \right)^2} = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \times \frac{49}{4} \lim_{x \to \pi} \frac{\sin^2 \left( \frac{7\left( x - \pi \right)}{2} \right)}{\frac{49}{4} \left( x - \pi \right)^2} = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \times \frac{49}{4} \lim_{x \to \pi} \frac{\sin^2 \left( \frac{7\left( x - \pi \right)}{2} \right)}{\left( \frac{7}{2}\left( x - \pi \right) \right)^2} = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \times \frac{49}{4} \lim_{x \to \pi} \left[ \frac{\sin\left( \frac{7\left( x - \pi \right)}{2} \right)}{\left( \frac{7}{2}\left( x - \pi \right) \right)} \right]^2 = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \times \frac{49}{4} \times 1 = f\left( \pi \right)\]
\[ \Rightarrow \frac{1}{5} \times \frac{49}{2} \times 1 = f\left( \pi \right)\]
\[ \Rightarrow \frac{49}{10} = f\left( \pi \right)\]

Hence, the given function will be continuous at

\[x = \pi\]

\[f\left( \pi \right) = \frac{49}{10}\] .