I would like to know what would be the best procedure to evaluate the limits of the following functions; some explanation would be appreciated:

$$\lim_{\theta\rightarrow -\infty}\dfrac{\cos\theta}{3\theta}$$

and

$$\lim_{\theta\rightarrow \infty}\dfrac{\sin2\theta}{\theta}$$

The first one I have found is zero and I would suppose this is so since the dominant function here is $\frac{1}{\theta}$ and $\cos\theta$ just oscillating between $1$ and $-1$.

You are perfectly right. This argument works in both cases.

Precisely, we have $$0\le \left|\frac{\cos\theta}{3\theta}\right|\le\frac1{|3\theta|}\ \to\ 0$$ as $\theta\to \pm\infty$.

From here you can use the squeeze theorem, or either, you can also use the very definition of limit. For any $\varepsilon>0$ choose $\theta_0:=1/(3\varepsilon)$, then for all $\theta>\theta_0$ we have $$\left|\frac{\cos\theta}{3\theta}-0\right|< \frac1{3\theta_0}=\varepsilon\ .$$

• The above is for $\theta\rightarrow \infty$ right? It would then be $\theta_0=-1/(3\epsilon)$ for $x\rightarrow -\infty$. Feb 19 '13 at 16:44
• Exactly. $\ \!\!$ Feb 19 '13 at 17:08

You can use the same reasoning for both of them. If the numerator is bounded and the denominator grows big without bound, then the quotient vanishes in the limit.

• I guess that is what @Berci explained above. Is there another technique other than using the Squeeze Theorem? Feb 19 '13 at 16:07
• You could do a proof by contradiction, but we must consider then 3 cases: (1) the limit of the quotient is finite but non-zero, (2) the limit of the quotient is infinite, (3) the limit of the quotient does not exist. In cases (1) and (2), we can use the growth of the denominator to conclude that the numerator grows without bound, contradicting our boundedness assumption. I don't see a clear path to contradiction in case (3) without using Squeeze Theorem or something like it. Is there a reason you'd like to avoid the Squeeze Theorem? Feb 19 '13 at 16:17
• No, just wondering. Just looking for options to explain to my students. The many more possible and different solutions I have the more options I will have to explain it to them. As my question was worded, I was going for the Squeeze Theorem but thought there might be other ways. Feb 19 '13 at 16:22