( 3^h - 1) / h as h->0 = ?, (Without using L'Hopital)

Taylor series of function f(h):  f(h) = f(0) + df/dh₀*h + (d²f/dh²)₀*h²/2! + ...

If f(h) = 3^h then df/dh = ln3*3^h,  d²f/dh² = (ln3)²*3^h, etc. 

So, noting that 3⁰ = 1, we have:  3^h = 1 + ln3*h + (ln3)²*h²/2! + ...

(the ... indicates terms involving higher powers of h)

Hence 3^h - 1 = ln3*h + (ln3)²*h²/2! + ...

(3^h - 1)/h = ln3 + (ln3)²*h/2! + ...

As h → 0 all but the first term on the right hand side go to 0, so in the limit we have

lim_h→0(3^h - 1)/h = ln3 

(much easier using l'Hopital's rule though!)

I'd like to see this one done too please :)

Also

As a slightly seperate issue...

Why is

\(\frac{d}{dh}\;3^h =\;3^h\;log3\)

By definition, the derivative of f(x) at x = a is

\(\displaystyle \lim_{h \rightarrow 0}=\frac{f(a+h)-f(a)}{h}\),

so,

\(\displaystyle \lim_{h \rightarrow 0}=\frac{3^{h}-1}{h}=\lim_{h \rightarrow 0}\frac{3^{0+h}-3^{0}}{h}\)

is the derivative of   \(\displaystyle f(x)=3^{x}\) at x = 0.

That will be equal to \(\displaystyle \ln(3)\).

Tiggsy