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

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Answer 1 · 2018-04-01T15:38:26

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

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

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

$= e^3$

Answer 2 · 2018-04-01T16:03:51

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

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]}