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What is the limit of # (lnx)^2/x# as x approaches #oo#?

Answer 1

#0#

#(log_ex)^2/x = ((log_e x)/sqrt(x))^2# so #lim_{x->oo}(log_ex)^2/x = (lim_{x->oo}(log_e x)/sqrt(x))^2# but #lim_{x->oo}(log_e x)/sqrt(x)=lim_{x->oo}(e^{log_e x})/e^{sqrt(x)} = lim_{x->oo}(x)/e^{sqrt(x)}#

Making now #sqrt(x) = y#

#lim_{x->oo}(x)/e^{sqrt(x)}equivlim_{y->oo}(y^2)/e^y = 0#

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Answer 2

Christopher Allers

The limit of (lnx)^2/x as x approaches infinity 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.

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