Question

If function $f$ is differentiable at $x=1$ and $f\left(1\right)=1$ and $f^{\prime }\left(1\right)=2$, then $\underset{x\to 1}{lim}\frac{f\left(x\right)-1}{x-1}=$_____

• A. $2$
• B. $-1$
• C. $1$
• D. $-2$

Answer 1

The correct option is A $2$

Now,

$\underset{x\to 1}{lim}\frac{f\left(x\right)-1}{x-1}$ $\left(\frac{0}{0}\right)$ form,

Now applying L'Hospital's rule we get,

$=\underset{x\to 1}{lim}\frac{f^{\prime }\left(x\right)}{1}$

$=f^{\prime }\left(1\right)$

$=2$.