Question

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

Answer 1

Since $\underset{x\to 1}{\text{lim}}(x^2-1)=(1{)}^2-1$

= 1-1=0

$\therefore \underset{x\to 1}{\text{lim}}\frac{f\left(x\right)-2}{x^2-1}$ = $\pi$ becomes $\underset{x\to 1}{\text{lim}}$ [f(x)-2]=0

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