Question:
If the function $1(x)$ satisfies $\lim _{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}=\pi$, evaluate $\lim _{x \rightarrow 1} f(x)$.
Solution:
$\lim _{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}=\pi$
$\Rightarrow \frac{\lim _{x \rightarrow 1}(f(x)-2)}{\lim _{x \rightarrow 1}\left(x^{2}-1\right)}=\pi$
$\Rightarrow \lim _{x \rightarrow 1}(f(x)-2)=\pi \lim _{x \rightarrow 1}\left(x^{2}-1\right)$
$\Rightarrow \lim _{x \rightarrow 1}(f(x)-2)=\pi\left(1^{2}-1\right)$
$\Rightarrow \lim _{x \rightarrow 1}(f(x)-2)=0$
$\Rightarrow \lim _{x \rightarrow 1} f(x)-\lim _{x \rightarrow 1} 2=0$
$\Rightarrow \lim _{x \rightarrow 1} f(x)-2=0$
$\therefore \lim _{x \rightarrow 1} f(x)=2$