We have $\lim\limits_{x\to 1}\frac{f(x)-2}{x^2-1}$ = π Since $\lim\limits_{x\to 1}$ $(x^2-1)$ = 0 ∴ For $\lim\limits_{x\to 1}$ $\frac{f(x)-2}{x^2-1}$ to exist, we must have $\lim\limits_{x\to 1}$ [f (x) - 2] = 0 [Since $\lim\limits_{x\to 1}$ (f (x) - 2) ≠ 0, then the given limit can’t exist] ⇒ $\lim\limits_{x\to 1}$ f (x) - 2 = 0 ⇒ $\lim↙{x→1} f (x) = 2.