Question

If $f\left(x\right)=\frac{1-\cos \left(7\right(x-\pi \left)\right)}{5(x-\pi {)}^2},$ for $x\ne \pi$ is continuous at $x=\pi ,$ find $f\left(\pi \right)$

Answer 1

$RHL={lim}_{h\to 0}f(\pi +h)={lim}_{h\to 0}\left(\frac{1-cos\left[7\right(\pi +h-\pi \left)\right]}{5(\pi +h-\pi {)}^2}\right)={lim}_{h\to 0}\left(\frac{1-cos7h}{5h^2}\right)={lim}_{h\to 0}\left(\frac{1-(1-2si{n}^2(\frac{7h}{2}\left)\right)}{5h^2}\right)={lim}_{h\to 0}\left(\frac{2si{n}^2\left(\frac{7h}{2}\right)}{5h^2}\right)={lim}_{h\to 0}\left(\frac{2si{n}^2\left(\frac{7h}{2}\right)(\frac{7h}{2}{)}^2}{(\frac{7h}{2}{)}^{2}5h^2}\right)={lim}_{h\to 0}{\left(\right(\frac{sin\left(\frac{7h}{2}\right)}{\left(\frac{7h}{2}\right)}\left)\right)}^2\left(\right(\frac{2\times 49}{5\times 4}\left)\right)=\left(\frac{49}{10}\right)=4.9$

$\therefore$for function to be continuous at $x=\pi LHL=RHL=f\left(\pi \right)\therefore f\left(\pi \right)=4.9$