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I'm studying for my Cal 2 exam, but I can't remember how to do the following question:

Find the limit, as n approaches infinity, of the sequence or prove that the limit does not exist: $$\lim_{n \to {\infty}}{{3 \cdot n^2 \cdot \cos(n \cdot \pi)} \over {\sqrt{1+4 \cdot n^4}}}$$

Please guide me in the right path. Thanks!

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  • $\begingroup$ What is your limit approaching? You might want to add that in. $\endgroup$ – Zach466920 Jun 22 '15 at 15:35
  • $\begingroup$ It doesn't say, but I'm assuminf Infinity $\endgroup$ – Aftab Hussain Jun 22 '15 at 15:36
  • $\begingroup$ where is $x$ in your formula? $\endgroup$ – Dr. Sonnhard Graubner Jun 22 '15 at 15:36
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    $\begingroup$ @AftabHussain I made another edit, tell me if that looks right. $\endgroup$ – Zach466920 Jun 22 '15 at 15:38
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    $\begingroup$ Perfect! Thank you for the edits Zach and Danimal $\endgroup$ – Aftab Hussain Jun 22 '15 at 15:40
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We can extract $2n^2$ from the denominator:

$$\lim_{n \to {\infty}}{{3 n^2 \cdot \cos(n \pi)} \over 2n^2 {\sqrt{\frac{1}{4n^4}+1}}}$$

Simplify the fraction:

$$\lim_{n \to {\infty}}{{3 \cdot \cos(n \pi)} \over 2 {\sqrt{\frac{1}{4n^4}+1}}}$$

As $n \to \infty$, we can see that $\frac{1}{4n^4} \to 0$, so that $\sqrt{\frac{1}{4n^4} + 1} \to \sqrt{0 + 1} = 1$, so the limit equals:

$$\lim_{n \to {\infty}}{\frac{3}{2} \cos(n \pi)}$$

Which does not exist, as the values for $\cos(n\pi)$ keep fluctuating between $1$ and $-1$ as you increment $n$, so the sequence will tend to $\frac{-3}{2}, \frac 32, \frac{-3}{2}, \frac 32, \dots$

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    $\begingroup$ tl;dr version: the "dominant powers" of numerator and denominator are the same, so they balance each other. But the remaining term oscillates, so the limit doesn't exist $\endgroup$ – MichaelChirico Jun 22 '15 at 15:52
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$$(1) \quad \lim_{n \to {\infty}}{{3 \cdot n^2 \cdot \cos(n \cdot \pi)} \over {\sqrt{1+4 \cdot n^4}}}$$

This limit is undefined...

Proof:

Let's divide both the numerator and denominator by $n^2$ we'll get...

$$(2) \quad \lim_{n \to {\infty}}{{3 \cdot \cos(n \cdot \pi)} \over {\sqrt{1/n^4+4}}}$$

Apply the quotient rule...

$$(3) \quad {\lim_{n \to {\infty}}{3 \cdot \cos(n \cdot \pi)} \over {\lim_{n \to {\infty}}\sqrt{1/n^4+4}}}$$

Ok, the denominator clearly goes to $4$, but the numerator is indeterminate. Keep in mind that cosine is periodic, but since we approach infinity, we can't define its value. The most we can say is that it's between $-1$ and $1$.

Thus, the limit is undefined.

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  • $\begingroup$ The final bit about $\cos$ is slightly inaccurate: what if the expression had been $\cos\left(n\pi + \frac{\pi (n+1)}{2n}\right)$? (Note that this is a sequence $\Bbb N \to \Bbb R$, not a function $\Bbb R \to \Bbb R$.) $\endgroup$ – Lynn Jun 22 '15 at 15:59
  • $\begingroup$ @Mauris But it's not that expression. It's just a vanilla cosine ;) $\endgroup$ – Zach466920 Jun 22 '15 at 16:04
  • $\begingroup$ Agree -- that's why I said "slightly"! What I'm a bit worried about is this: "(something) is periodic, but since we approach infinity, we can't define its value" is reasoning that applies here, but it sounds a lot like it might generalize, while there are certainly cases where it doesn't apply (see my example.) $\endgroup$ – Lynn Jun 22 '15 at 18:06
  • $\begingroup$ @Mauris well, now they have your post to warn them :) $\endgroup$ – Zach466920 Jun 22 '15 at 19:32

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