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

Reuben Brennan

Answered question

2022-03-30

Evaluate the following limits.

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

delai59qk

Beginner2022-03-31Added 8 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 q 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 L'Hospital rule,

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

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

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

= lim_{n \to \infty} n^{2} \cdot 0

= 0

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?