Solve the limit as h approaches 0. LIMIT BELOW!

Question

anonymous · Answer

$$\lim_{h \rightarrow 0}\frac{ 7^{h}-1 }{ h }$$

anonymous · Answer

I found out the answer if supposed to be like log(7) or ln(7) according to WolframAlpha. Cant see how to get there really. I thought maybe i'd multiply by ln/ln , and get $$\frac{ h-\ln 1 }{ \ln h }$$    but that doesn't work because you can't take the natural log of 0 on the bottom still

anonymous · Answer

this is the derivative of $7^x$ at $x=0$ 
the derivative of $7^x$ is $7^x\ln(7)$ so at $x=0$ you get $7^0\ln(7)=\ln(7)$

anonymous · Answer

i don't think there is a way to do this directly if you have not gotten to derivatives yet. you can do it numerically by trying smaller and smaller values of $h$ and approximating it with a decimal

anonymous · Answer

I haven't really learned derivatives yet.. i just know i was suposed to transform that limit so that I could solve it

anonymous · Answer

you could also use l'hopital, but that presupposes knowing derivatives

anonymous · Answer

oh ok. well thanks

anonymous · Answer

you might try rewriting as 
$$\lim_{n\to \infty}\frac{7^{\frac{1}{n}}-1}{\frac{1}{n}}$$  but i am not sure what that is going to buy you.

anonymous · Answer

well thanks anyway. I guess i can just ask my teacher about it tomorrow