Answer:
Explanation:
[tex]\lim_{x \to \ 0} \frac{2e^x-2}{x}[/tex]
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 [tex]\lim_{x\to \ c} \frac{f'(x)}{g'(x)}[/tex]
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 [tex]\lim_{x \to \ 0} 2e^x[/tex]
Plugging in zero we see this gives us 2 as 2(e^0) = 2(1) = 2.
Hence, [tex]\lim_{x \to \ 0} \frac{2e^x-2}{x}[/tex] = 2
