Question

If the function$f\left(x\right)$ satisfies $\underset{x\to 1}{lim}\frac{f\left(x\right)-2}{x^2-1}=\pi$ evaluate $\underset{x\to 1}{lim}f\left(x\right)$

Answer 1

$\underset{x\to 1}{lim}\frac{f\left(x\right)-2}{x^2-1}=\pi$

$\Rightarrow \frac{{lim}_{x\to 1}(f\left(x\right)-2)}{{lim}_{x\to 1}(x^2-1)}=\pi$

$\Rightarrow \underset{x\to 1}{lim}(f\left(x\right)-2)=\pi \underset{x\to 1}{lim}(x^2-1)$

$\Rightarrow \underset{x\to 1}{lim}(f\left(x\right)-2)=\pi (1^2-1)$

$\Rightarrow \underset{x\to 1}{lim}(f\left(x\right)-2)=0$

$\Rightarrow \underset{x\to 1}{lim}f\left(x\right)-\underset{x\to 1}{lim}2=0$

$\Rightarrow \underset{x\to 1}{lim}f\left(x\right)-2=0$
$\therefore \underset{x\to 1}{lim}f\left(x\right)=2$