Finding a limit of a quotient with trig 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!
 A: 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 $\frac{-3}{2}, \frac 32, \frac{-3}{2}, \frac 32, \dots$
A: $$(1) \quad \lim_{n \to {\infty}}{{3 \cdot n^2 \cdot \cos(n \cdot \pi)} \over {\sqrt{1+4 \cdot n^4}}}$$
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.
