# How to find $\lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$ and $\lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$

How can I determine these limits:

$$a) \lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$$

$$b) \lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$$

Note I cannot use l'Hospital. But I know that $\lim_{n\rightarrow 0} \frac{\sin n}{n} = 1$ and $\lim_{n\rightarrow 0} \frac{\cos n -1}{n} = 0$

I already found the limit of $\lim_{n\rightarrow 0} \frac{\sin(3n) \sin(2n)}{n^2}$:

$$\lim_{n\rightarrow 0} \frac{\sin(3n) \sin(2n)}{n^2} = \lim_{n\rightarrow 0} \frac{\sin(3n)}{n} \lim_{n\rightarrow 0} \frac{ \sin(2n)}{n} = 3 \lim_{n\rightarrow 0} \frac{\sin(3n)}{3n} 2\lim_{n\rightarrow 0} \frac{ \sin(2n)}{2n} = 3 \cdot 2 = 6$$

Can I use a similar approach here?

• Notice: $\lvert\cos\alpha\rvert\leqslant1$ First, $$\cos\alpha\leqslant1\implies-\color{red}{\cos\alpha+1\geqslant0}$$ $$\frac{n}{\sin(2n)-\cos\left(\frac{n}{2}\right)+1}\leqslant\frac{n}{\sin(2n)}\to\frac{1}{2}$$ – Invisible Apr 19 at 12:29

Hints: Your approaches are quite correct. Just explore them further.

For (a), try rewriting the interior expression as $\pi\cdot \dfrac{\sin n}{n} \cdot \cos n$.

For (b), divide top and bottom by $n$ and write it as $\dfrac{1}{\frac{\sin 2n}{n} - \frac{\cos \frac{n}{2} - 1}{n}}$

• Thanks so much! Thanks to your help, I was able to find the limits myself. Very helpful that you didn't give a complete solution right away, it's much better to work things out by yourself :) – IronMan12 May 29 '14 at 14:30

Hint For part (a) you could also use the double angle formula in reverse

\begin{align} \frac{\pi}{n} \sin n \cos n = \pi\frac{\sin 2n}{2n} \end{align}

Which you already know the limit to.

Hint Part (b) can be done by dividing the numerator and denominator by n.

\begin{align} (a)\quad\lim_{n\rightarrow 0} \cos\left(\frac{\pi\sin n \cos n}{n}\right)&=\cos\left(\lim_{n\rightarrow 0}\frac{\pi\sin n \cos n}{n}\right)\\ &=\cos\left(\lim_{n\rightarrow 0}\frac{\pi\sin n \cos n}{n}\right)\\ &=\cos\left(\pi\lim_{n\rightarrow 0}\cos n\frac{\sin n}{n}\right)\\ &=\cos(\pi)\\ &=-1. \end{align} Note that, it is valid to move the limit since cosine function is continuous. \begin{align} (b)\quad\lim_{n\rightarrow 0} \dfrac{\dfrac nn}{\dfrac{2\sin2n}{2n} -\left(\dfrac{ \cos\frac{n}{2}-1}{2\cdot\frac{n}{2}}\right)}=\frac{1}{2}. \end{align}

For (a), you can use a very similar approach. Since $\cos(x)$ is continuous on $\mathbb{R}$, you can pull the limit inside, and you can rewrite the limit as: \begin{align*} \lim_{n\to 0} \cos\left(\frac{\pi}{n}\sin(n)\cos(n)\right) &= \lim_{n\to 0} \cos\left(\frac{\sin(n)}{n}\pi\cos(n) \right)\\ &= \cos\left(\lim_{n\to 0} \frac{\sin(n)}{n} \pi \cos(n) \right) \end{align*}

For (b), try \begin{align*} \lim_{n\to 0} \frac{n}{\sin(2n) - \cos\left(\frac{n}{2}\right) +1} &= \lim_{n\to 0} \frac{1}{\frac{\sin(2n)}{n} - \frac{\cos\left(\frac{n}{2}\right) -1}{n}}\\ &= \lim_{n\to 0} \frac{1}{2\frac{\sin(2n)}{2n} - \frac{1}{2}\frac{\cos\left(\frac{n}{2}\right) -1}{\frac{n}{2}}} \end{align*}

(You will want to make sure you understand the steps in these equalities. The first is dividing top and bottom by $n$, but the second is manipulating each term in the denominator to make them look like the limits you already know.)

And from here, your previous approach will prove useful.