Question

Let $L=\underset{x\to 0}{lim}\frac{x^2-1+e^x}{2x}$,If L is finite,then

• A. $L=\frac{1}{2}$
• B. $L=1$
• C. $L=\frac{1}{4}$
• D. $L=\frac{1}{32}$

Answer 1

The correct option is A $L=\frac{1}{2}$

We have,

$L=\underset{x\to 0}{lim}\frac{x^2-1+e^x}{2x}$

This is in the form of $\frac{0}{0}$

So, using L -Hospital rule ,

$\Rightarrow L=\underset{x\to 0}{lim}\frac{2x+e^x}{2}=\frac{1}{2}$