Question

Help me to find it please? I am studying for my midterm in real analysis and I got problem in it.

Answer 1

To find the limit without l'Hospital's Rules, use `lim_(xrarroo)(1+1/x)^x = e`

Explanation:

`lim_(xrarroo)((x+1)/x)^(3x) = lim_(xrarroo)(1+1/x)^(3x)`

`= (lim_(xrarroo)(1+1/x)^x)^3`

`= e^3`

Answer 2

`lim_(x->oo) ((x+1)/x)^(3x) = e^3`

Explanation:

Write the function as:

`((x+1)/x)^(3x) = (e^ln((x+1)/x))^(3x) = e^(3x ln((x+1)/x)`

Consider now the limit:

`lim_(x->oo) 3x ln((x+1)/x)`

This is in the indeterminate form `0*oo` but we can reduce it to the form `0/0` by writing as:

`lim_(x->oo) 3x ln((x+1)/x) = 3 lim_(x->oo) ln((x+1)/x) /(1/x)`

and the use l'Hospital's rule:

`lim_(x->oo) 3x ln((x+1)/x) = 3 lim_(x->oo) (d/dx ln((x+1)/x)) /(d/dx ( 1/x))`

`lim_(x->oo) 3x ln((x+1)/x) = 3 lim_(x->oo) (x/(x+1) (x-(x+1))/x^2) /(-1/x^2)`

`lim_(x->oo) 3x ln((x+1)/x) = 3 lim_(x->oo) x/(x+1) (-1/x^2) /(-1/x^2)`

`lim_(x->oo) 3x ln((x+1)/x) = 3 lim_(x->oo) x/(x+1) = 3`

As `e^x` is a continuous function we then have:

`lim_(x->oo) e^(3x ln((x+1)/x) )= e^((lim_(x->oo) 3x ln((x+1)/x) )) = e^3`

graph{ ((x+1)/x)^(3x) [-1, 100, -2, 49.2]}