Question

Extending the functions in col. 1 as ${lim}_{x\to \pi }f\left(x\right)$

Answer 1

A) ${lim}_{x\to \pi }f\left(x\right)={lim}_{x\to \pi }\frac{1-\cos \left(7\right(x-\pi \left)\right)}{x-\pi }={lim}_{x\to \pi }\frac{2{\sin }^2\left(\right(7/2\left)\right(\pi -x\left)\right)}{\left(\right(7/2\left)\right(\pi -x){)}^2}.\frac{49}{4}(\pi -x)=\left(2\right)\left(1\right)\left(\frac{49}{4}\right(\pi -\pi \left)\right)=0$
$\Rightarrow f\left(\pi \right)=0$
B) ${lim}_{x\to \pi }f\left(x\right)={lim}_{x\to \pi }\frac{1-\cos \left(7\right(x-\pi \left)\right)}{(x-\pi {)}^2}={lim}_{x\to \pi }\frac{2{\sin }^2\left(\right(7/2\left)\right(\pi -x\left)\right)}{\left(\right(7/2\left)\right(\pi -x){)}^2}.\frac{49}{4}=\left(2\right)\left(1\right)\left(\frac{49}{4}\right)=\frac{49}{2}$
$\Rightarrow f\left(\pi \right)=\frac{49}{2}$
C) ${lim}_{x\to \pi }f\left(x\right)={lim}_{x\to \pi }\frac{\sin x}{1-x^2/{\pi }^2}={lim}_{x\to \pi }\frac{\sin (\pi -x)}{(\pi -x)(\pi +x)}.{\pi }^2=\frac{{\pi }^2}{2\pi }=\frac{\pi }{2}$
$\Rightarrow f\left(\pi \right)=\frac{\pi }{2}$
D) ${lim}_{x\to \pi }f\left(x\right)={lim}_{x\to \pi }\frac{\sin 7x}{\sin 2x}={lim}_{u\to 0}\frac{\sin 7(\pi -u)}{\sin 2(\pi -u)}={lim}_{u\to 0}\frac{-\sin 7u}{\sin 2u}=\frac{-7}{2}$
$\Rightarrow f\left(\pi \right)=\frac{-7}{2}$