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

Asked by Pragya Singh | 1 year ago |  81

1 Answer

Solution :-

\( \lim\limits_{x \to 1}\dfrac{f(x)-2}{x^2-1}=\pi\)

\(\dfrac{\lim\limits_{x \to 1}f(x)-2}{\lim\limits_{x \to 1}(x^2-1)}=\pi\)

\( \lim\limits_{x \to 1}(f(x)-2)=\pi\lim\limits_{x \to 1}(x^2-1)\)

\( \lim\limits_{x \to 1}(f(x)-2)=\pi (1^2-1)\)

\( \lim\limits_{x \to 1}(f(x)-2)=0\)

\( \lim\limits_{x \to 1}(f(x)- \lim\limits_{x \to 1}2=0\)

\( \lim\limits_{x \to 1}(f(x)- 2=0\)

\( \lim\limits_{x \to 1}(f(x)=2\)

Answered by Abhisek | 1 year ago

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