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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives

If the function $f(x)$ satisfies $ \lim\limits_{x \to 1} \large\frac{f(x)-2}{x^2-1}= \pi$ , evaluate $ \lim\limits_{x \to 1} f(x)$

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$ \lim\limits_{x \to 1} \large\frac{f(x)-2}{x^2-1}= \pi$
$ \Rightarrow \large\frac{\lim\limits_{x \to 1}(f(x)-2)}{\lim\limits_{x \to 1}(x^2-1)}= \pi$
$ \Rightarrow \lim\limits_{x \to 1}(f(x)-2) = \pi \lim\limits_{x \to 1}(x^2-1)$
$ \Rightarrow \lim\limits_{x \to 1}(f(x)-2) = \pi (1^2-1)$
$ \Rightarrow \lim\limits_{x \to 1}(f(x)-2) = 0$
$ \Rightarrow \lim\limits_{x \to 1} f(x)- \lim\limits_{x \to 1} 2=0$
$ \Rightarrow \lim\limits_{x \to 1} f(x)- 2=0$
$ \therefore \lim\limits_{x \to 1} = 2$
answered Apr 7, 2014 by thanvigandhi_1
 

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