We have an answer to "Evaluate the following limits. limn→∞n2ln⁡(nsin⁡1n)"

aortiH

Answered question

2020-11-16

Evaluate the following limits.

lim_{n \to \infty} n^{2} \ln (n \sin \frac{1}{n})

funblogC

Skilled2020-11-17Added 91 answers

Consider the provided expression,

lim_{n \to \infty} n^{2} \ln (n \sin \frac{1}{n})

Evaluate the following limits.
We can write as,

lim_{n \to \infty} n^{2} \ln (n \sin \frac{1}{n}) = lim_{n \to \infty} n^{2} \cdot lim_{n \to \infty} \ln (n \sin \frac{1}{n})

Simplifying further,
Apply the chain rule of the limit.

lim_{n \to \infty} n^{2} \ln (n \sin \frac{1}{n}) = lim_{n \to \infty} n^{2} \cdot (lim_{n \to \infty} \ln (n \sin \frac{1}{n}))

Apply the algebraic property,

a \cdot b = \frac{a}{\frac{1}{b}}, b \neq 0.

\sin (\frac{1}{n}) n = \frac{\sin (\frac{1}{a})}{\frac{1}{a}}

Now, the expression becomes.

lim_{n \to \infty} n^{2} \ln (n \sin \frac{1}{n}) = lim_{n \to \infty} n^{2} \cdot (lim_{n \to \infty} \ln (\frac{\sin (\frac{1}{a})}{\frac{1}{a}}))

Apply LHospital

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