What is the limit of $(x/(x+1))^x$ as x approaches infinity?

Question

64539 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-21T20:55:02

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

First, we will use the following:

$e^ln(x) = x$
Because $e^x$ is continuous on $(-oo,oo)$ , we have $lim_(x->oo)e^f(x) = e^(lim_(x->oo)f(x))$

With these:

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

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

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

Next, we will use L'Hopital's rule:

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

$=lim_(x->oo)(d/dxln(x/(x+1)))/(d/dx1/x)$

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

$lim_(x->oo)-x/(x+1)$

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

$=-1$

Putting them together, we get our final result.

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

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