Can someone show me how to evaluate lim x->0+ (1+X)^(1/x) using L'Hopital's rule? I know I need to convert it into proper form, a fraction I think, to solve it. I think it has something to do with logarithms but I can't figure it out. Thanks!

Expert Answers

An illustration of the letter 'A' in a speech bubbles

You need to take the logarithm of the limit such that:

`ln lim _(x->0) (1+x)^(1/x) = lim_(x->0) ln ((1+x)^(1/x))`

Using logarithmic identity yields:

`lim_(x->0) ln ((1+x)^(1/x)) = lim_(x->0) (1/x)ln (1+x)`

`lim_(x->0) (1/x)ln (1+x) = lim_(x->0) (ln (1+x))/x = (ln 1)/0 = 0/0`

You may use l'Hospital's theorem such that:

`lim_(x->0) (ln (1+x))/x = lim_(x->0) ((ln (1+x))')/(x') `

`lim_(x->0) ((ln (1+x))')/(x') = lim_(x->0) (1/(1+x))/1`

Substituting 0 for x yields:

`lim_(x->0) ((ln (1+x))')/(x') = 1/(1+0) = 1`

Hence, evaluating the limit logarithm of the limit yields:

`ln lim _(x->0) (1+x)^(1/x) = 1 => lim _(x->0) (1+x)^(1/x) = e^1`

`lim _(x->0) (1+x)^(1/x) = e`

Hence, evaluating the limit, using logarithm and l'Hospital's theorem, yields `lim _(x->0) (1+x)^(1/x) = e` .

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial Team