Question

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

Answer 1

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

Explanation:

Ok this requires a few wee tricks. We want to find

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

Because the exponential and natural log functions are inverse to each other they cancel out so we can rewrite this as

`lim_(xrarroo) exp(ln((x+1)/(x))^x)`

Using rules of logs we can bring the exponent down:

`lim_(xrarroo) exp(xln((x+1)/(x)))`

Now notice that the bit that actually changes is the exponent of the exponential function, so that's what we focus on:

`exp(lim_(xrarroo) (xln((x+1)/(x))))`

If at all possible we want to use L'hopital's rule on the limit. For this we need it to be in indeterminate form, eg `0/0 or (oo)/(oo) etc..`

Rewrite as:

`exp(lim_(xrarroo) (ln((x+1)/(x))/(1/x)))`

Check the numerator and denominator separately:

`lim_(xrarroo) 1/x = 0`

`lim_(xrarroo) ln((x+1)/x) = lim_(xrarroo) ln((1+1/x)/1) = ln(1) = 0`

so we have `0/0` and are in the correct form for L'hopital's.

`exp(lim_(xrarroo) ((d/(dx)(ln((x+1)/(x))))/(d/(dx)(1/x))))`

Computing the derivative of the numerator using chain and quotient rules:

`d/(dx)(ln((x+1)/x)) = x/(x+1)*(-1)/(x^2) = - 1/(x(x+1))`

For the denominator, easiest just to rewrite and use power rule

`d/(dx)( x^(-1)) = -1/x^2`

We now have:

`exp(lim_(xrarroo)((- 1/(x(x+1)))/(-1/x^2))) = exp(lim_(xrarroo)(x/(x+1)))`

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

So we end up with

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