Finding Limit: Hi, I've Got 2 Questions

[kemmy]

Hi, I've got two questions here that I'm stuck on.
The first:

Homework Equations

Find the Limit as x->0 of [(e^4x)-1-4x]/x^2

The Attempt at a Solution

So far:
I got that the Limit as x->0 is indeterminate form 0/0 so I tried L'Hopital's:

to find the Limit as x->0 of [(e^4x)-4]/2x

but that gave me a Limit of -3/0 . So I think that means that there's no limit. But the book insists that the Limit is 8. and I have no idea how it got there.


the second question I'm confused about is

Homework Equations

Find the Limit as x->2 of (x+2)/[(x-2)^4] , and explain why L'Hopital's Rule does not apply.

The Attempt at a Solution

so
The Lim x->2 = 4/0 so it's not 0/0 indeterminate so L'Hopital's isn't applicable.

But after that I'm a bit stuck on how to get the Limit. I know that because x can't equal 0 in the denominator that x cannot equal 2. So there's a vertical asymptote there.
So...I know I can look at it from each side
The Limit as x->0- and as x->0+ .
So-
The Limit as x->0+=infinity
The Limit as x->0-=infinity.
is that right? and is their an easier way to find that then just plugging in numbers near x=2?

-Thanks, any help would be much appreciated!

 

[nicksauce]

The derivative of e^(4x) is 4e^(4x)... correct this and it should work out.