Question

If $f\left(x\right)=\frac{sin(e^{x-2}-1)}{log(x-1)}$, then $\underset{x\to 2}{lim}f\left(x\right)$ is given by

• A. -2
• B. -1
• C. 0
• D. 1

Answer 1

The correct option is D 1

$f\left(x\right)=\underset{x\to 2}{lim}\frac{\sin (e^{x-2}-1)}{\log (x-1)}$

This is $\frac{0}{0}$ form so by using L Hospital rule , differentiating the numerator and denomirator

$\Rightarrow f\left(x\right)=\underset{x\to 2}{lim}\frac{\cos (e^{x-2}-1)\left(e^{x-2}\right)}{\left(\frac{1}{x-1}\right)}\Rightarrow f\left(x\right)=\underset{x\to 2}{lim}\cos (e^{x-2}-1)\left(e^{x-2}\right)(x-1)\Rightarrow f\left(x\right)=\cos (e^{2-2}-1)\left(e^{2-2}\right)(2-1)\Rightarrow f\left(x\right)=1$

So option $D$ is correct.