Proof check

[Darya]

I tried to recall the standard proof of [MATH]\lim_{x \to 0} \frac{e^x-1}{x}=1[/MATH] but then I got an idea and did this:
[MATH]\lim_{x \to 0} \frac{e^x-1}{x}=\lim_{x \to 0} \frac{((1+x)^{1/x})^x-1}{x}=\lim_{x \to 0} \frac{1+x-1}{x}=1[/MATH]I just wanted to check if it's valid to prove it this way. Thanks a lot!

 

[Steven G]

Yes, it is OK but is missing some steps. You replaced e^x with ((1+x)^1/x)^x but they are not equal. Do you see that?

Try to do it again.

 

[Darya]

Yes, it is OK but is missing some steps. You replaced e^x with ((1+x)^1/x)^x but they are not equal. Do you see that?

Try to do it again.

Yes, e is equal to the limit of (1+x)^(1/x) when x approaches 0. The whole expression is in terms of limit so I skipped this step but agree that it should've been written in the proof.