Question

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

Answer 1

`Lt_(h->o)(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 `Lt_(h->o)(e^h-1)/h=f(0)=1`

Answer 2

See below

Explanation:

If this is part of the derivation of the fact that `d/dx ( e^x) = 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

`\lim _{h\to \infty } (1+ 1/h )^h = e`

or, here
`\lim _{h\to 0 } (1+ h )^(1/h) = e`

So

`lim_(h to 0) ((e^h)-1)/(h)`

`= lim_(h to 0) (((1+ h )^(1/h))^h-1)/(h)`

`=lim_(h to 0) (1+ h - 1)/(h)`

`= 1`