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

Question

Christopher Allers · Accepted Answer

#Lt_(h->o)(e^h-1)/h=1# 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#

Luke Alvarez · Answer

See below

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 o \infty } (1+ 1/h )^h = e# or, here #\lim _{h o 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#

Cameron Alexander · Answer

undefined To find the limit as h approaches 0 for the expression ((e^h)-1)/(h), we can use L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately and then evaluate the limit again.

Differentiating the numerator, we get e^h. 
Differentiating the denominator, we get 1.

Now, we can evaluate the limit again by substituting h = 0 into the differentiated expression:

lim(h→0) (e^h)/(1)

Simplifying further, we have:

lim(h→0) e^h

Finally, substituting h = 0 into the expression, we get:

lim(h→0) e^0 = e^0 = 1

Therefore, the limit as h approaches 0 for ((e^h)-1)/(h) is equal to 1.