First principles differentiation question

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SUMMARY

The derivative of e^{2x} can be obtained from first principles using the limit definition of the derivative. The expression simplifies to e^{2x}(\frac{e^{2h}-1}{h}) as h approaches 0. By substituting u = 2h, the limit can be expressed as 2\frac{e^u - 1}{u}, which approaches 2 as u approaches 0. Thus, the derivative of e^{2x} is confirmed to be 2e^{2x}.

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Homework Statement


Obtain the derivative of [tex]e^{2x}[/tex] from first principles.


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The Attempt at a Solution


I am upto the point where i have [tex]e^{2x}(\frac{e^{2h}-1}{h})[/tex] Where LIM h->0.

I just don't see how 2 is obtained from what's in the ().
 
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[tex]\frac{e^{2h}-1}{h}= 2\frac{e^{2h}-1}{2h}= 2\frac{e^u -1}{u}[/tex]
if you let u= 2x.

Now, do you know what the limit, as u goes to 0 of
[tex]\frac{e^u- 1}{u}[/tex]
is?
 
It tends to 1? I think i understand this now, thanks very much for your help.
 

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