Calculate Compound limit

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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=+∞