# Calculate Compound limit

I am in great difficulty to calculate this limit. $$\lim_{x\to0}(\cos x) \frac{1}{x\sin(2x)}$$ I know that the $$\lim\limits_{x\to0} \cos x=0$$, and the fraction it will be $$\frac{1}{0}$$ but it doesn't exist. So I don't know how to solve it, what kind of mistake am I making. Is here someone who may explain it to me, or tell me what rules I have to watch? Also, I have tied Comparison theorem, but it will be $$- 1/1\leq x\sin x(2x)\leq 1/1$$ and this is also incorrect.

• It is $\lim_{x\to 0}\cos(x)=1$ Sep 6, 2019 at 17:03
• what do you understand with lim x->0 (cos(x)) cos(x)->0 so x->pi/2? Sep 6, 2019 at 17:04
• The limit is of the form $\frac10$, so it doesn't exist. Why is that a problem? Sep 6, 2019 at 17:06
• @saulspatz nope. It's moltiplicate. (COSX) (1/xsinx(2x))
– Ciao
Sep 6, 2019 at 17:07
• Do you not know that $a\left({1\over b}\right)={a\over b}$? Sep 6, 2019 at 17:27

Write $$\frac{\cos(x)}{x\cdot 2\sin(x)\cos(x)}$$

• I can't understand how you get x*2sin(x)cos(x)? Cos I try to solve it again, and I have cos(x)(x sin (2x)) and it will be 0*0. So at the end I have 1/0=doesnt exist.
– Ciao
Sep 6, 2019 at 18:10
• It is $$\sin(2x)=2\sin(x)\cos(x)$$ Sep 6, 2019 at 18:11

For small enough $$|x|$$ we have $$x\sin 2x<2x^2$$therefore $${\cos x\over x\sin 2x}>{{1\over 2}\over 2x^2}={1\over 4x^2}$$

$$\sin x\sim x$$ in the process $$x\to 0$$, thus $$\lim _{x\to 0} \frac{\cos x}{x\sin 2x} = \lim _{x\to 0} \frac{1}{2x\sin x} = \lim _{x\to 0}\frac{1}{2x^2} = +\infty.$$

For the purposes of this problem I'm going to assume that $$cos(x)$$ is defined as the standard cosine function with $$\lim_{x\rightarrow0}\cos(x)$$=1 and $$\frac{\sqrt{2}}{2}<\cos(x)<1$$ for $$x\in[-\frac{\pi}{4},\frac{\pi}{4}]$$. I will also assume that $$\sin(x)$$ is defined as the standard sine function with $$\sin(x)=-\sin(-x)$$ and $$0\leq\sin(x)\leq1$$ for $$x\in[0,\frac{\pi}{2}]$$ and $$\sin(0)=0$$.

I contend that for any $$\epsilon$$ you give me I can find a $$\delta$$ such that $$|x|<\delta \implies \frac{\cos(x)}{x \sin(2x)}>\epsilon.$$

We know that for $$\delta <\pi/4$$:

1. $$\frac{\cos(x)}{x \sin(2x)} > 0$$ since $$\cos(x)>0$$ and since sign($$x$$) = sign($$\sin(2x)$$) then $$x\sin(2x)>0$$
2. $$\frac{\cos(x)}{x \sin(2x)}>\frac{1}{|x|\sqrt{2}}$$

Therefore choose $$\delta = \min(1,\frac{1}{\sqrt{2}\epsilon})$$

This establishes that the limit increases without stop as $$x$$ approaches 0.

• Did you mean $\lim\limits_{x\to0}\cos x=\color{red}1?$ Sep 6, 2019 at 18:30
• @J.W.Tanner I did. Thank you for spotting that Sep 6, 2019 at 18:32

I solved it in this way (1)/( x sin x(2x))=1/0= ∞. So I'm going to calculate right and left limit and I have + ∞ and -  ∞. But if I'm looking that (1) /(2x^2(sin x) /x) =1/2x^2=+∞