Evaluate the following limits. \(\displaystyle\lim_{{{n}\rightarrow\infty}}{n}^{{2}}{\ln{{\left({n}{\sin{{\frac{{{1}}}{{{n}}}}}}\right)}}}\)

Reuben Brennan 2022-03-30 Answered
Evaluate the following limits.
limnn2ln(nsin1n)
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Answers (1)

delai59qk
Answered 2022-03-31 Author has 8 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, bq0.
sin(1n)n=sin(1a)1a
Now, the expression becomes.
limnn2ln(nsin1n)=limnn2(limnln(sin(1a)1a))
Apply L'Hospital rule,
limnn2ln(nsin1n)=limnn2(limnlnlimn(cos(1a)a21a2))
=limnn2(limnlnlimn(cos(1n)))
=limnn2(limnln1))
=limnn20
=0
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