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31. If the function f(x) satisfies $\lim_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \pi$ , evaluate $\lim_{x \rightarrow 1} f (x)$

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Given

$\lim_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \pi$

Now,

$\lim_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \frac{\lim_{x \rightarrow 1}(f (x)-2)}{\lim_{x \rightarrow 1}(x^2-1)}=\pi$

${\lim_{x \rightarrow 1}(f (x)-2)}=\pi{\lim_{x \rightarrow 1}(x^2-1)}$

${\lim_{x \rightarrow 1}(f (x)-2)}=\pi{(1-1)}$

${\lim_{x \rightarrow 1}(f (x)-2)}=0$

${\lim_{x \rightarrow 1}f (x)}=2$

Posted by

Pankaj Sanodiya

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