Question

If the function $f\left(x\right)$ satisfies $\underset{x\to 1}{lim}\frac{f\left(x\right)-2}{x^2-1}=\pi$, then $\underset{x\to 1}{lim}f\left(x\right)=$

• A. $2$
• B. $3$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

$2$

It is given that, $\underset{x\to 1}{lim}\frac{f\left(x\right)-2}{x^2-1}=\pi$ exists and equal to $1$.

Clearly the denominator is zero at $x=1$, so for above limit to exist

numerator should also be zero, as x approaches $1$.

Therefore, $\underset{x\to 1}{lim}f\left(x\right)=2$

Note: If $\underset{x\to 1}{lim}f\left(x\right)\ne 1$, given limit will not exist