Question

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

Answer 1

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

Explanation:

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)))`

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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`

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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`