How to find the exact value of

Question

lim (2^h - 1)/ h
h->0

Thanks in advance.

If you use
lim dx->0 of [f(a + dx) - f(a)]/dx = f'(a)
and recognize 1=2<sup0.
Then use f(x)=2^x
Thus lim dx->0 [2^0+dx - 2<sup0]/dx = f'(0).
This means find d/dx 2^x and evaluate at x=0.

You can think of the dx as delta x, or increment of x.
I'm not sure if you've used L'Hospital's rule or not, but that's another way you could solve this problem too.