Question

What is the limit of `(e^x + x)^(1/x)` as x approaches infinity?

Answer 1

`e`

Explanation:

`(e^x+x)^(1/x)=e(1+x/e^x)^(1/x)` now making `y=x/e^x` we have
`e(1+y)^(e^(-x)/y)=e((1+y)^(1/y))^(e^(-x))` here
`{(x->oo),(y(x)->0):}` then

`lim_(x->oo)(e^x+x)^(1/x)=e lim_(x->oo)(lim_(y->0)(1+y)^(1/y))^(e^(-x))=e cdot e^0=e`

Answer 2

`lim_(xrarroo)(e^x+x)^(1/x)=e`

Explanation:

`L=lim_(xrarroo)(e^x+x)^(1/x)`

A helpful tool in finding limits with exponential functions like this is to use the equivalence `a=e^lna`. Thus

`L=lim_(xrarroo)e^(ln(e^x+x)^(1/x))`

This can be rewritten using the log rule `log(a^b)=bloga` as

`L=lim_(xrarroo)e^(1/xln(e^x+x))`

We can move the limit inside of the exponential function:

`L=e^(lim_(xrarroo)(ln(e^x+x)/x)`

Notice that `lim_(xrarroo)ln(e^x+x)/x` is in the indeterminate form `oo/oo`, so L'Hopital's rule applies. We can take the derivatives of the numerator and denominator:

`L=e^(lim_(xrarroo)((d/dxln(e^x+x))/(d/dxx))`

`L=e^(lim_(xrarroo)((e^x+1)/(e^x+x))/1`

`L=e^(lim_(xrarroo)(e^x+1)/(e^x+x)`

You may immediately recognize that this limit is `1`, since the `e^x` terms in the numerator and denominator will overpower the other terms, so as `x` approaches infinity, `(e^x+1)/(e^x+x)approxe^x/e^x=1`, but in case this isn't enough for you we can do L'Hopital's again since we are in the `oo/oo` form.

`L=e^(lim_(xrarroo)(e^x/(e^x+1))`

And L'Hopital's once more, as we're still in `oo/oo`:

`L=e^(lim_(xrarroo)(e^x/e^x)`

`L=e^(lim_(xrarroo)1)`

`L=e^1`

`L=e`