Question

How do you evaluate `(1+a/x)^x` as x approaches infinity?

Answer 1

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

Explanation:

Define `y=lim_(x->oo)(1+a/x)^x`. Now take the natural log to get `ln(y)=lim_(x->oo)x*ln(1+a/x)`. Now ignore the left side and focus on the right side. The right side can be rewritten as

`lim_(x->oo)(ln(1+a/x))/(1/x)`

You can try evaluating this limit by plugging in infinity directly. You'll get `0/0` which is indeterminate form. So apply L'Hôpital's rule and differentiate the numerator and denominator.

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

So we have evaluated this limit, but remember this limit is equal to `ln(y)`. So solve for y to get the final answer.

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

Answer 2

`lim_{x->infty}(1+a/x)^x=e^a`

Explanation:

`(1+a/x)^x = ((1+a/x)^{x/a})^a`
`lim_{x->infty}(1+a/x)^x = (lim_{x->infty}(1+a/x)^{x/a})^a = e^a`