Algebraic Method for Finding the Limit of (2^h - 1)/h as h Approaches 0

[Miike012]

How do you differentiate lim (2^h - 1)/h ?
h --> 0
I would like to know how to do it algebraically instead of picking values of x that approach 0 and plugging it into 2^x.

 

[LCKurtz]

How do you differentiate lim (2^h - 1)/h ?
h --> 0
I would like to know how to do it algebraically instead of picking values of x that approach 0 and plugging it into 2^x.

I presume you mean how do you evaluate that limit. If you know the derivative of e^x you can do it this way:

$$\lim_{h\rightarrow 0}\frac{2^h-1}{h}=<br /> \lim_{h\rightarrow 0}\frac{e^{h\ln 2}-1}{h}=<br /> (\ln 2)\ \lim_{h\rightarrow 0}\frac{e^{h\ln 2}-1}{h\ln 2}$$

Not let t = h ln(2) giving

$$(\ln 2)\ \lim_{t\rightarrow 0}\frac{e^{t}-1}{t}$$

That gives ln(2) as the answer because the limit of the fraction is 1. You can see that either by recognizing that difference quotient as the derivative of e^x at x = 0 or by applying L'Hospital's rule to it if you have had that.