Limit of (e^tanx - e^x)/(tanx-x) as x tends to 0 is

I can't even evaluate it by using l'hospital's rule...Please write a detailed answer!!

Thanks to anyone who answers:)

lim(x-->0) (e^tanx - e^x)/(tanx-x)=
=lim(x-->0) eˣ * lim(x-->0)(e^(tanx-x) - 1)/(tanx-x)=
= eº * lim(x-->0)(e^(tanx-x) - 1)/(tanx-x)=
= lim(x-->0)(e^(tanx-x) - 1)/(tanx-x)=
Let v=tanx-x
= lim(v-->0)(e^v - 1)/v=
Let u=e^v - 1, v=ln(u+1)
= lim(u-->0) u/ln(u+1)=
= lim(u-->0) 1/ln[(u+1)^(1/u)]=
= 1/ln[e]= 1

http://www.wolframalpha.com/input/?i=lim…

Well the answer is one, as can be shown from a plot. I'm stumped how to solve it analytically, but once a limit is found you can prove it using that epsilon delta thing. Been a while since I did that....

Maybe the squeeze theorem can help you find the limit analytically, but I believe the most practical way (this might be considered a cheat by a mathematician but in physics we do this all of the time) is the expand the function as a series. It expands to:

1+x+x^2/2+x^3/3+x^4/24+19x^5/120+.....…

Since we are talking about when x is very small, close to zero, we can ignore terms of second order or higher (a very small number squared is an even smaller number, and a very small number to the fifth power is so tiny we don't care about its contribution), so if we just say

(e^tanx-e^x)/(tanx-x)=1+x (approximately), then.....

lim x->0 of 1+x is 1

The simple answer is 1 using the online limit calculator provided in my source.

For detailed workout, see similar questions:

http://uk.answers.yahoo.com/question/ind…

http://answers.yahoo.com/question/index?…

The first answer provides a link to an article explaining the technique.