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

2 Answers
Shwetank Mauria · Jim H
Sep 14, 2016

#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#

Eddie
Sep 14, 2016

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#

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