Question #261f3

1 Answer
sente
Nov 28, 2016

#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#

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