But there's so many terms

Calculus Level 3

$\large \lim_{n\to\infty} \frac{n^k \sin^2(n!) }{n+2}$

Fix variable $0 < k < 1$ . What is the above limit?

1 -1 Does not exist 0

2 solutions

Calvin Lin Staff
Nov 9, 2016

Since $0 < k < 1$ , we know that

$\lim_{n\rightarrow \infty} \frac{ n^k } { n+2 } = 0$

Since the limit of the above 0, and we know that $0 \leq \sin^2 (n!) \leq 1$ , and we can multiply this term in to conclude that

$\lim_{n\rightarrow \infty} \frac{ n^k \sin^2 (n!) } { n+2 } = 0$

@Josh Banister @Star Light This is one way to approach the problem.

Just because the numerator increases, doesn't mean that the limit is infinite. We still have to account for the (rate of) increase in the denominator.

Calvin Lin Staff - 4 years, 7 months ago

Samarth Sangam
Aug 28, 2014

$since\quad 0<k<1\\ lim_{ n\rightarrow \infty }{ { n }^{ k } }=0\\ therefore\quad the\quad answer\quad is\quad 0!$

I did the same,but didn't saw A! 😢

Akhil Bansal - 5 years, 9 months ago

Let k = 1/2. It is an increasing function so as n get's large, n^k get larger too.

Josh Banister - 5 years, 11 months ago

I agree that the solution does not explain how to solve this problem.

However, the answer is indeed 0.

Calvin Lin Staff - 4 years, 7 months ago

$\frac{64}{16} = \frac{\not64}{1\not6} = \frac{1}{4}$

Josh Banister - 4 years, 7 months ago

I agree with Josh. Shouldn't n^k increase as n increases.

Star Light - 4 years, 7 months ago

But there's so many terms

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