Question
If the function f(x) satisfies $\mathop {\lim }\limits_{x \to 1} \frac{{f(x) - 2}}{{{x^2} - 1}} = \pi ,$ then evaluate $\mathop {\lim }\limits_{x \to 1} f(x)$.
✓
Answer
Given, $\mathop {\lim }\limits_{x \to 1} \frac{{f(x) - 2}}{{{x^2} - 1}} = \pi \Rightarrow \frac{{\mathop {\lim }\limits_{x \to 1} [f(x) - 2]}}{{\mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)}} = \pi $
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} [f(x) - 2] = \pi \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) - 2 = \pi \left( {{1^2} - 1} \right)$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) - 2 = \pi \times 0 \Rightarrow \mathop {\lim }\limits_{x \to 1} f(x) - 2 = 0$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) = 2$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} [f(x) - 2] = \pi \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right)$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) - 2 = \pi \left( {{1^2} - 1} \right)$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) - 2 = \pi \times 0 \Rightarrow \mathop {\lim }\limits_{x \to 1} f(x) - 2 = 0$
$ \Rightarrow \quad \mathop {\lim }\limits_{x \to 1} f(x) = 2$
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