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

Answer 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#

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Answer 2

Ezekiel Alexander

To find the limit of x/(ln(1+2e^x)) as x approaches infinity, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (1)/(1+2e^x). As x approaches infinity, e^x also approaches infinity, so the limit becomes 1/(1+2∞), which simplifies to 1/∞. Therefore, the limit is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.