Help with understanding of L'Hospitals Rule

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SUMMARY

The limit of the function (lnx)^2/x as x approaches infinity is evaluated using L'Hôpital's Rule. Initially, both the numerator and denominator approach infinity, leading to the application of L'Hôpital's Rule, resulting in the limit of 2lnx/x. A second application of L'Hôpital's Rule simplifies this to 2/x, which definitively approaches 0 as x approaches infinity. The conclusion is that the limit of (lnx)^2/x is 0.

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  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's Rule
  • Knowledge of logarithmic functions
  • Basic differentiation techniques
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  • Learn about indeterminate forms in calculus
  • Explore the behavior of logarithmic functions at infinity
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Students in calculus courses, educators teaching limits and derivatives, and anyone seeking to deepen their understanding of L'Hôpital's Rule and its applications.

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Homework Statement


This was a question from our lecture notes, just not sure how the prof arrived at the answer.

lim x->infinity (lnx)^2/x

Homework Equations




lim x->infinity (lnx)^2/x
lim x->infinity 2lnx/x

The Attempt at a Solution




so both the numerator and denominator are going towards infinity, and by L'H it the lim x->infinity 2lnx/x
so this means that the numerator is 'growing' faster than the denominator, a constant x? also, how does one arrive at the conclusion that the limit is 0?
 
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(lnx)^2/x = (2/x)(ln x)

derivative of ln x = 1/x

Go from there. I'm drunk.
 
shocklightnin said:

Homework Statement


This was a question from our lecture notes, just not sure how the prof arrived at the answer.

lim x->infinity (lnx)^2/x

Homework Equations




lim x->infinity (lnx)^2/x
lim x->infinity 2lnx/x

The Attempt at a Solution




so both the numerator and denominator are going towards infinity, and by L'H it the lim x->infinity 2lnx/x
so this means that the numerator is 'growing' faster than the denominator, a constant x? also, how does one arrive at the conclusion that the limit is 0?

Just apply L'Hospital's rule again since you are still in an indeterminate inf/inf:
\frac{2}{x}

It should now be pretty sensible that it approaches 0 as x approaches infinity.
 
RoshanBBQ, thanks! Completely slipped my mind that sometimes we have to apply L'H more than once.
 

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