Question

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

Answer 1

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

Explanation:

Notice that

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

and if the limit exists,

`lim_{x -> oo} ( e^(x ln(1 + 4/x)) ) = e^{lim_{x -> oo}(x ln(1+4/x))}`

as the exponential function is continuous everywhere.

To evaluate the limit at the exponent, we first write it as

`x ln(1 + 4/x) = frac{ln(1 + 4/x)}{1/x}`

Since the form is indeterminate `0/0`, use the L'hospital rule.

`lim_{x->oo}(ln(1+4/x)/(1/x)) = lim_{x->oo}(frac{frac{d}{dx}(ln(1+4/x))}{frac{d}{dx}(1/x)})`

`= lim_{x->oo}(frac{-4/x^2}{(1+4/x)}/(-1/x^2))`

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

`= frac{4}{1+0}`

`= 4`

Therefore, the limit is `e^4`.

Answer 2

If you are familiar with the sometimes definition of `e`, as `lim_(urarroo)(1+1/u)^u = e`, then we don't need l"Hospital.

Explanation:

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

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

Now, with `u = x/4`, we have

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

`= (lim_(urarroo)(1+1/u)^u)^4 = e^4`