Question

Question #261f3

Answer 1

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

Explanation:

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

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

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

The above step is valid due to the continuity of `f(x)=e^x`.

Solving the limit in the exponent, we have

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

`=lim_(x->oo)(d/dxln((x+1)/(x+3)))/(d/dx1/x)`

The above step applies L'Hopital's rule to a `0/0` indeterminate form.

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

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

`=lim_(x->oo)(2/(x^2+4x+3))/(-1/x^2)`

`=lim_(x->oo)-(2x^2)/(x^2+4x+3)`

`=lim_(x->oo)-2/(1+4/x+3/x^2)`

`=-2/(1+0+0)`

`=-2`

Substituting this into our original work, we get the final result:

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