Question

I need help finding the derivative of e^2x.
lim (x→0)〖(e^2x-1)/x〗:

Answer 1

Chain rule, man. 2e^(2x). But are you thinking of applying l'Hopital's rule? Because it doesn't work here. The limits of top and bottom don't evaluate equally.

Answer 2

Yes, I was trying to figure out if this limit would produce an indeterminate form to see if L'Hopital's rule could apply, but I didn't know how to find the derivative of e^2x.

Answer 3

Oh, my bad! They do!

Answer 4

Forgot the -1 at the side!

Answer 5

What is the answer to the problem?

Answer 6

2, if I did math right.

Answer 7

Can you show me the work so I can see how you came up with 2?

Answer 8

campbell_st seems to be on the case, and he's better with LaTeX than I am, so I'll defer to him. ;)

Answer 9

divide top and bottom by x^2

then the problem becomes

\[\lim_{x \rightarrow0} (e^(2x -1)/x = \lim_{x \rightarrow 0} xe^(2x -1) \]

which means the limit is zero...

Answer 10

What? The derivative of 2x-1 is 2, not x.

Answer 11

Check yo work, campbell. ಠ_ಠ

Answer 12

Also, the -1 is outside of the exponent, otherwise l'Hopital's rule can't be applied.

Answer 13

Proof: http://www.wolframalpha.com/input/?i=lim+as+x+approaches+0+%28e^%282x%29-1%29%2F%28x%29

Answer 14

welll look at it parts...then x(e^2x - 1) so its the limit of the parts

e^2x = 1 as x approaches zero

x = 0 as x approches zero

so its 0(1 -1) still zero

Answer 15

Well, as campbell and I disagree, I will post my solution. With corroboration from WolframAlpha. http://www.wolframalpha.com/input/?i=lim+as+x+approaches+0+%28e^%282x%29-1%29%2F%28x%29

Now, the current limit that we take is in the indeterminate 0/0 form. (e^(2x)-1)/x. Remember that the -1 must be outside, and not inside, the exponent, otherwise the form cannot have l'Hopital's applied.

Taking derivative of top and bottom nets us 2e^(2x). The derivative of e^(n) is (dn/dx)e^(n), so it's 2e^(2x). Evaluating THIS as x approaches 0 nets us 2*1 or just 2.

Yo campbell, check your work.