Question

Limit as x approaches 0 of (2e^x-2)/x

Answer 1

Answer:

Explanation:

$\lim_{x \to \ 0} \frac{2e^x-2}{x}$

So first, we have to plug in zero and see if we can evaluate this limit simply from that.

When we plug in zero we get: (2e^0-2)/0

e^0 is 1 so we have 2-2/0 or 0/0. So we have an indeterminate form type 0/0.

This means we have to apply L'Hospital's Rule.

As a reminder L'Hospitals Rule is $\lim_{x\to \ c} \frac{f'(x)}{g'(x)}$

Meaning that we take the derivative of the top and bottom function as the approach some value "c". We can do this with a 0/0 indeterminate form.

So:

The derivative of 2e^x - 2 is just 2e^x

and the derivative of x is 1

So we are left with $\lim_{x \to \ 0} 2e^x$

Plugging in zero we see this gives us 2 as 2(e^0) = 2(1) = 2.

Hence, $\lim_{x \to \ 0} \frac{2e^x-2}{x}$ = 2