limit of (e^x-1)/X as x approachs zero?

Question

I just started AP Calculus BC this year and it really bugs me when i am given rules with no explaination. So yesterday i figured out proofs for many rules like the Power rule for finding derivatives. So i was thinking this morning and i had used my answer for the derivative of y=lnx and y= e^x in eachother. To find the derivative of e^x i assumed the derivative of lnx was 1/x and to find the dirivative of lnx i assumed the derivative of e^x was itself. which i didnt like very much because its kind of like an endless loop of assumptions. So, i saw that you could find the derivative of e^x using the simple definition of a derivative. but that required the assumption that the limit of (e^x-1)/x as x approaches zero is one, but i could not proof this assumption algabraicly only by graphing. So if someone could please show me how to prove this, or simply show me another way to prove my two original derivatives without using the answers for eachother in both of them. you can use the answer

of one in the other but not in both. i will show my work to clarify. also i dont know proper notations with dy and dx and such so some of it may be off but the core math is sound

y=lnx
e^y=x

dy e^y= dx 1
dy/dx=1/e^y
dy/dx= 1/x

y=e^x
lny=lne^x
lny=x
dy 1/y = dx 1
dy/dx= y
dy/dx = e^x

or

e^x using the definition of a derivative

lim [e^(x+h)-e^x]/h
x->0
lim (e^xe^h-e^x)/h
x->0
lim e^x(e^h-1)/h
x->0
now there is my problem i am assuming based on the graph(my calc book shows me a graph and makes an allusion to some complicated calc for the rest) so can someone please show me why e^h-1/h is 1 algabraiclly. sorry for the spelling errors.

thank you DG i dont really understand the whole binomal expansion yet as i have only been in calculus for two weeks. i just enjoy righting proofs for the rules they give me, and doing other problems such as dy/dx of y=x^x^x^x. But anyway thank you for the answer but is there anyway a beginer in calc could get it i tryed with the definition of a derivative and l'hopitals but i kept running in circles and i am trying to do this without assuming e^x is itelf in the problem since that is what i am trying to prove with this equation. Anyway thank you :-)

ok so far DG has the best answer only i dont know enough claculus yet to get how to pull it out into a binomial expansion. Yea,i geuss the answers are limited by my dearth of knowledge in math... But with l'hospital you end up with

d/dx (e^x-1)/d/dx (x)
d/dx e^(x+h)-1-(e^x-1)/h/1(which is d/dx for x)
e^xe^h-e^x/h (ones cancel)
e^x(e^h-1/h)
so i am essentially right back where i started with out making the assumption that it is equal to one.
If i did any of that wrong could some some please write out the entire process to go thru using l'hospitals with no assumptions or "this is this because its an established rule of calculus" such as e^x-1/X is one

and i can't use the sum or difference rule which is what it looks like you did in your l'hospitals rule because i can't assume the dirivative of e^x is e^x because that is what im trying to prove. i can't use the solution to my proble to try and solve my problem. The solution to my question may simply be i dont know enough calculus yet to satisfy what i am looking for and that DG has the only correct answer im looking for.

Answer 1

The binomial expansion of e^x is
e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...

so e^x - 1 = x + x^2/2! + x^3/3! + x^4/4! + ...

This divided by x has only the first term free of x = 1
All others terms have x in them

so as x tends to zero, the limit of (e^x - 1)/x is 1

So much for the limit

If u want to prove the derivative, just take the derivativeof the xpansion and u get the same value as the expansion is infinte

so d/dx (e^x) = e^x

Answer 2

1

Answer 3

L'Hospital rules apply as of a zero/zero condition is present.

Taking the derivative wrt x of both the numerator and denominator

f(x) = (e^x -1) /x
L'Hospital = d/dx(e^x-1) = e^x
L'Hospital = d/dx(x) = 1

Expression is now lim(x go to zero) of e^x/1 = 1/1 or 1

Hope this explanation helps.

Answer 4

Use L'Hospital's Rule: derivative of top divided by derivative of bottom.
(e^x-1)/x -> d(e^x-1)/d(x) -> e^x/1 as x goes to 0 => e^0/1 = 1/1 =1