Evaluate the following limits. lim_{nrightarrowinfty}n^2ln(nsinfrac{1}{n})

aortiH 2020-11-16 Answered
Evaluate the following limits.
limnn2ln(nsin1n)
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Expert Answer

funblogC
Answered 2020-11-17 Author has 91 answers
Consider the provided expression,
limnn2ln(nsin1n)
Evaluate the following limits.
We can write as,
limnn2ln(nsin1n)=limnn2limnln(nsin1n)
Simplifying further,
Apply the chain rule of the limit.
limnn2ln(nsin1n)=limnn2(limnln(nsin1n))
Apply the algebraic property, ab=a1b, b0.
sin(1n)n=sin(1a)1a
Now, the expression becomes.
limnn2ln(nsin1n)=limnn2(limnln(sin(1a)1a))
Apply LHospital
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