How do you find the limit of #x/(ln(1+2e^x))# as x approaches infinity?

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Answer 1 · 2016-04-27T19:21:45

1

#lim_(x->oo) x/(ln(1+2e^x)) = oo/ln(1+2e^oo)=oo/oo#

This is an indeterminate type so use l'Hopital's Rule. That is, find the limit of the derivative of the top divided by the derivative of the bottom.

#=lim_(x->oo) 1/(1/(1+2e^x) xx 2e^x)#

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

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

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

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

#=1/(2e^oo)+1#

#=0+1#

#=1#