Given ((e^h)-1)/(h) how do you find the limit as h approaches 0?

Sep 14, 2016

$L {t}_{h \to o} \frac{{e}^{h} - 1}{h} = 1$

Explanation:

We use series expansion of ${e}^{x}$ as

e^x=1+x+1/(2!)x^2+1/(3!)x^3+1/(4!)x^4+.......

Hence e^h=1+h+1/(2!)h^2+1/(3!)h^3+1/(4!)h^4+.......

and e^h-1=h+1/(2!)h^2+1/(3!)h^3+1/(4!)h^4+....... or

(e^h-1)/h=f(h)=1+1/(2!)h+1/(3!)h^2+1/(4!)h^3+.......

Hence $L {t}_{h \to o} \frac{{e}^{h} - 1}{h} = f \left(0\right) = 1$

Sep 14, 2016

See below

Explanation:

If this is part of the derivation of the fact that $\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$, then clearly you cannot use L'Hopital's Rule or a Taylor Expansion, as you would be assuming the actual outcome in the derivation

I may be wrong but I think we should still be allowed to use Bernoulli's compounding formula

$\setminus {\lim}_{h \setminus \to \setminus \infty} {\left(1 + \frac{1}{h}\right)}^{h} = e$

or, here
$\setminus {\lim}_{h \setminus \to 0} {\left(1 + h\right)}^{\frac{1}{h}} = e$

So

${\lim}_{h \to 0} \frac{\left({e}^{h}\right) - 1}{h}$

$= {\lim}_{h \to 0} \frac{{\left({\left(1 + h\right)}^{\frac{1}{h}}\right)}^{h} - 1}{h}$

$= {\lim}_{h \to 0} \frac{1 + h - 1}{h}$

$= 1$