# 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...

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!

### 1 Answer | Add Yours

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` **.