How to solve limits for trigonometric sequences? $\lim_{n \to \infty} n^3\left(1-\cos\left(\frac{1}{n}\right)\right)\sin\left(\frac{1}{n}\right)$

What kind of criteria do I have to use to solve for trigonometric limits.

For Example those kind of problems:

$$\lim_{n \to \infty} n^3\left(1-\cos\left(\frac{1}{n}\right)\right)\sin\left(\frac{1}{n}\right)$$

or

$$\lim_{n \to \infty} \frac{\sin(n+2)\sin(n-2)}{\cos(n+2)\cos(n-2)}$$

I obviously can't get there with the sandwich-theorem or the ratio test.

Thanks for help community.

• There is always the chance to apply L' Hospitals rule. That wold work for the upper limit. However one does not learn too much using L'Hospital, that's why I am kinda against this method. Maybe there is another way.. For the lower limit, you should ask, whether the limit is well defined. – Imago Dec 21 '15 at 6:36
• Multiply rop and bottom by $1+\cos(1/n)$ and let $1/n=t$. Now it's nearly over. – André Nicolas Dec 21 '15 at 6:37
• So I am allowed to use hopiâl for sequences as well? Thnx – Fill Dec 21 '15 at 6:43

Notice, use double angle formula: $\cos 2A=1-2\sin^2 A$, $$\lim_{n\to \infty}n^3\left(1-\cos\left(\frac{1}{n}\right)\right)\sin\left(\frac{1}{n}\right)$$ $$=\lim_{n\to \infty}n^3\left(1-1+2\sin^2\left(\frac{1}{2n}\right)\right)\sin\left(\frac{1}{n}\right)$$ $$=\lim_{n\to \infty}\frac{n}{2}\left(\frac{\sin^2\left(\frac{1}{2n}\right)}{\left(\frac{1}{2n}\right)^2}\right)\sin\left(\frac{1}{n}\right)$$ $$=\frac{1}{2}\lim_{n\to \infty}\left(\frac{\sin\left(\frac{1}{2n}\right)}{\left(\frac{1}{2n}\right)}\right)^2\lim_{n\to \infty}\frac{\sin\left(\frac{1}{n}\right)}{\left(\frac{1}{n}\right)}$$ $$=\frac{1}{2}(1)^2(1)=\color{red}{\frac 12}$$

• Hi, thanks a lot. I don't see how you let the $n^3$ go $n/2$ though? – Fill Dec 21 '15 at 7:06
• Alright, notice in third step $n^3$ changes to $\frac{\sin^2\left(\frac{1}{2n}\right)}{\left(\frac{1}{2n}\right)^2}$ & $n/2$ is left in numerator – Harish Chandra Rajpoot Dec 21 '15 at 7:09
• Which trigonometric equality to you use? $x^3 = ?$ I'm sorry I still don't see it. I appreciate your patience. – Fill Dec 21 '15 at 7:13
• I have already shown the trig identity of double angle to use is $\cos 2A=1-2\sin^2 A$ – Harish Chandra Rajpoot Dec 21 '15 at 7:15
• Yeah I see that, I don't see why this would impact $n^3$ – Fill Dec 21 '15 at 7:17

For the first one, set $1/n=2h$ to get

$$\lim_{h\to0}\dfrac{(1-\cos2h)\sin2h}{(2h)^3}=\dfrac12\cdot\left(\lim_{h\to0}\dfrac{\sin h}h\right)^2\cdot\lim_{h\to0}\dfrac{\sin2h}{2h}$$

• I don't see where this is going yet. Why are you taking the limit to 0 and not to infinity? – Fill Dec 21 '15 at 6:43
• because $1/n=2h$ so that when $n \to \infty$, so does $h \to 0$ – chandresh Dec 21 '15 at 6:52
• Gotcha, thanks. How do you get from the first step to the second one? – Fill Dec 21 '15 at 7:03
• @Fill, Use $$\cos2A=1-2\sin^2A$$ – lab bhattacharjee Dec 21 '15 at 7:34

You can use series approximations. For example,

$$1 - \cos x = 1 - \left(1 - \frac {x^2}{2} + O(x^4)\right)$$

and $\sin x = x + O(x^3)$. Applying this to the first limit gives

$$n^3 \left(1 - \cos \frac 1 n\right) \sin \frac 1 n = n^3 \left(\frac 1 {2n^2} + O\left(\frac{1}{n^4}\right)\right)\left(\frac 1 n + O\left(\frac 1 {n^3}\right)\right) = \frac 1 2 + O\left(\frac 1 n\right)$$

• So I can use tayloring to solve for limits of sequences? I didnt know that this would hold. Thanks! – Fill Dec 21 '15 at 6:40