Question

Question #6c442

Answer 1

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

Explanation:

Write the function as:

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

We evaluate now:

`lim_(x->oo) xln(1+1/x)`

This is in the indeterminate form `0*oo`, but we can transform it to the form `0/0` and solve it using l'Hospital's rule:

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

`lim_(x->oo) xln(1+1/x) = lim_(x->oo) (d/dx ln(1+1/x))/(d/dx (1/x))`

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

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

Now, as `e^x` is a continuous function we have:

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

So, in conclusion:

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