Question

What is the limit of `(1+4/x)^x` as x approaches infinity?

Answer 1

`y=e^4`

Explanation:

`lim_(x->oo) (1+4/x)^x = (1+4/oo)^oo = 1^ oo`

This is an indeterminate type so we use l'Hopital's Rule. That is, find the limit of the derivative of the top divided by the derivative of the bottom.

Let `y=(1+4/x)^x`

`lny=ln (1+4/x)^x`-> Take `ln` of both sides

`lny=xln (1+4/x)`

`lny=ln (1+4/x)/x^-1`

`lim_(x->oo) lny = lim_(x->oo) ln (1+4/x)/x^-1`

`lim_(x->oo) lny =lim_(x->oo) (1/(1+4/x) *-4/x^2)/(-1/x^2)`

`lim_(x->oo) lny=lim_(x->oo) 1/(1+4/x) *-4/x^2 xx -x^2`

`lim_(x->oo) lny=lim_(x->oo) 4/(1+4/x)`

`lny= 4/(1+4/oo)` Note that the limit of ln y is just ln y since it is a constant

`lny=4/(1+0)=4/1=4`

`e^lny=e^4`

`y=e^4`

Answer 2

`e^4`

Explanation:

Use `lim_(urarroo)(1+1/u)^u = e`

`(1+4/x)^x = ((1+1/(x/4))^(x/4))^4`

As `xrarroo`, the ration `x/4rarroo` so the inside expression goes to `e`.

`lim_(urarroo) ((1+1/(x/4))^(x/4))^4= (lim_(urarroo) (1+1/(x/4))^(x/4))^4`

`= (e)^4 = e^4`