Evaluating limit as h->0 of (e^h-1)/h

[autodidude]

I'm trying to differentiate e^x from first principles but I can't find a way to manipulate this expression \frac{e^h-1}{h} so I can evaluate the limit without getting 0/0

 

[eumyang]

Use the definition of e:
lim_{h \rightarrow 0} \left( 1 + h \right)^{1/h} = e
So for small values of h,
e \approx \left( 1 + h \right)^{1/h}, or
e^h \approx 1 + h.

Replace e^h in \frac{e^h-1}{h} with 1 + h and go on from there.

 

[Diffy]

L'hopitals rule works here too :-)

http://en.wikipedia.org/wiki/L'Hôpital's_rule

 

[Mute]

L'hopitals rule works here too :-)

http://en.wikipedia.org/wiki/L'Hôpital's_rule

In order to use L'Hopital's rule, you need to know the derivative of e^x, but that's what the OP is trying to prove. To use L'Hopital's rule in this case would be circular logic.

 

[eumyang]

L'hopitals rule works here too :-)

http://en.wikipedia.org/wiki/L'Hôpital's_rule

I'm not sure how. The OP is trying to "differentiate e^x from first principles," which I assume means that he/she want to find the derivative of e^x by using the limit definition:
\frac{d}{dx}e^x = lim_{h \rightarrow 0} \frac{e^{x+h} - e^x}{h} = ...
I don't think you can use L'Hopitals' Rule when we "don't know" the derivative of e^x yet.EDIT: Beaten to it by Mute.

 

[Diffy]

Oops my bad. I didn't read.

 

[autodidude]

Use the definition of e:
lim_{h \rightarrow 0} \left( 1 + h \right)^{1/h} = e
So for small values of h,
e \approx \left( 1 + h \right)^{1/h}, or
e^h \approx 1 + h.

Replace e^h in \frac{e^h-1}{h} with 1 + h and go on from there.

Don't we get 0/0 again?

Limit h->0

\frac{1+h-1}{h}

 

[eumyang]

Don't we get 0/0 again?

Limit h->0

\frac{1+h-1}{h}

Um, you need to simplify the expression.
lim_{h \rightarrow 0}\frac{1+h-1}{h} = lim_{h \rightarrow 0} \frac{h}{h} = ...

 

[autodidude]

^ Ah, thanks a lot...must've had a brain meltdown :p