Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If the function $f(x)$ satisfies $\displaystyle\lim _{x \rightarrow 1} \frac{f(x)-2}{x^2-1}=\pi$

Limits and Derivatives

Solution:

Given, $ \displaystyle\lim _{x \rightarrow 1} \frac{f(x)-2}{x^2-1}=\pi$
$\Rightarrow \frac{\displaystyle\lim _{x \rightarrow 1}[f(x)-2]}{\displaystyle\lim _{x \rightarrow 1}\left(x^2-1\right)}=\pi$
$\Rightarrow \displaystyle\lim _{x \rightarrow 1}[f(x)-2]=\pi \displaystyle\lim _{x \rightarrow 1}\left(x^2-1\right)$
$\Rightarrow \displaystyle\lim _{x \rightarrow 1} f(x)-2=\pi\left(1^2-1\right)$
$\Rightarrow \displaystyle\lim _{x \rightarrow 1} f(x)-2=\pi \times 0$
$\Rightarrow \displaystyle\lim _{x \rightarrow 1} f(x)-2=0$
$\Rightarrow \displaystyle\lim _{x \rightarrow 1} f(x)=2$