evaluate lim_h\to\0 e^{2(x+h)}-e^{2x}..

[wind]

Hi i am having trouble with this question, on my exam review

evaluate

\(\displaystyle \L\mbox{\lim_{h\to\0}\frac{e^{2(x+h)}-e^{2x}}{h}\)

\(\displaystyle \L\mbox{\lim_{h\to\0}\frac{lne^{2(x+h)}-lne^{2x}}{h}\)

\(\displaystyle \L\mbox{\lim_{h\to\0}\frac{(2x+2h)lne-(2x)lne}{h}\)

\(\displaystyle \L\mbox{\lim_{h\to\0}\frac{(2x+2h)-(2x)}{h}\)

\(\displaystyle \L\mbox{\lim_{h\to\0}\frac{2h}{h}\)

\(\displaystyle \L\mbox{\lim_{h\to\0}\ 2\)

but the answer is supposed to be \(\displaystyle \L\ 2e^{x}\)

pleas help

 

[pka]

Come on! That asks you to see that is the limit is the derivative of \(\displaystyle e^{2 x}\).

 

[skeeter]

here's a big clue ... you do not need to evaluate the limit, but you do need to recognize the form of the limit. Compare to this limit which should be burned in your memory ...

\(\displaystyle \L f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\)

\(\displaystyle \L f'(x) = \lim_{h \rightarrow 0} \frac{e^{2(x+h)} - e^{2x}}{h}\)

do you "see" what f(x) is in your limit? ... the limit is f'(x) = 2e^2x

 

[wind]

ok thanks, I did'nt think of that .

 

[soroban]

Hello, wind!

Big blunder in your first step . . .

Evaluate: \(\displaystyle \L\:\lim_{h\to\0}\frac{e^{2(x+h)}\,-\,e^{2x}}{h}\)

\(\displaystyle \L\lim_{h\to\0}\frac{\ln\left[e^{2(x+h)}\right]\,-\,\ln\left[e^{2x}\right]}{h}\;\;\) . . . What did you do here?

You took the log of the numerator (only) . . . how? .why?

Besides, \(\displaystyle \ln(A\,-\,B)\) is not equal to \(\displaystyle \ln(A)\,-\,\ln(B)\)