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

Given, $\underset{x\to 1}{lim}\frac{f\left(x\right)-2}{x^2-1}=\pi$
$\Rightarrow \frac{\underset{x\to 1}{lim}\left(f\right(x)-2)}{\underset{x\to 1}{lim}(x^2-1)}=\pi$
$\Rightarrow \underset{x\to 1}{lim}\left(f\right(x)-2)=\pi \times \underset{x\to 1}{lim}(x^2-1)$
$\Rightarrow \underset{x\to 1}{lim}f\left(x\right)-\underset{x\to 1}{lim}2=\pi \times (\underset{x\to 1}{lim}{x}^2-\underset{x\to 1}{lim}1)$
$\Rightarrow \underset{x\to 1}{lim}f\left(x\right)-2=\pi \times (1^2-1)$
$\Rightarrow \underset{x\to 1}{lim}f\left(x\right)-2=\pi \times 0$
$\Rightarrow \underset{x\to 1}{lim}f\left(x\right)-2=0$
$\Rightarrow \underset{x\to 1}{lim}f\left(x\right)=2$

Thus, $\underset{x\to 1}{lim}f\left(x\right)=2$