How to find the following limit? $ \lim_{x \to 0} \frac{x}{\sin(2x)\cos(3x)}$ I am stuck with this limit problem
$$\lim_{x \to 0} \frac{x}{\sin(2x)\cos(3x)} $$
Any hints are appreciated. Also, I can't use L'Hopital's.
 A: We know that: $\cos(3x)\to1$ as $x\to0$, so the only difficulty you're left with is to prove that: $$\lim_{x\to0}\dfrac{x}{\sin(2x)}=\dfrac12$$ and as a hint you can use: $$\lim_{x\to0}\dfrac{{x}}{\sin(2x)}=\lim_{x\to0}\dfrac12\dfrac{2{x}}{\sin(2x)}\quad\color{grey}{\sf and}\quad\lim_{x\to0}\dfrac{x}{\sin(x)}=1.$$
A: Consider this.
$$ \lim_{x \rightarrow 0}\,\dfrac{x}{\sin\,(2x)\cos\,(3x)} = \left( \lim_{x \rightarrow 0}\,\dfrac{x}{\sin\,(2x)} \right) \left( \lim_{x \rightarrow 0}\,\dfrac{1}{\cos\,(3x)} \right) $$
As $\cos\,(3x) \rightarrow 0$ when $x \rightarrow 0$, then $\lim_{x \rightarrow 0}\,\frac{1}{\cos\,(3x)} = 1$, and you can apply L'Hoptial's Rule to find the other limit.
A: You know that
$$ \lim_{x\rightarrow 0 }\frac{\sin(x)}{x} = \lim_{h \rightarrow 0}\frac{\sin(0+h) - \sin(0)}{h} = \sin'(0) = \cos(0) = 1.$$
Furthermore, $\cos(3x) \rightarrow \cos(0) = 1$ for $x\rightarrow 0$. This should be enough of a hint.
A: Using trigonometric identities, you can prove that
$$\sin{2x}\cos{3x}=2\sin{(x)}\cos^2{(x)}(2\cos{(2x)}-1).$$
Then using the rule that the limit of a product is the product of the limits, 
$$\lim_{x\to 0}\frac{x}{\sin{2x}\cos{3x}}=\lim_{x\to 0}\frac{x}{2\sin{(x)}\cos^2{(x)}(2\cos{(2x)}-1)}\\
=\frac12\left(\lim_{x\to 0}\frac{x}{\sin{(x)}}\right)\left(\lim_{x\to 0}\frac{1}{\cos^2{(x)}(2\cos{(2x)}-1)}\right)\\
=\frac12 (1) (\frac{1}{1\cdot(2-1)})=\frac12.$$
A: You may take advantage of the fact that $\sin{x} \to x$ as $x \to 0$ so, therefore, your limit becomes:
$$ L  = \lim_{x \to 0} \frac{x}{\sin{2x}  \,\cos{3x}} = \lim_{x \to 0} \frac{x}{2 x \, \cos{3x}}  = \lim_{x \to 0} \frac{1}{2 \, \cos{3x}}  = \frac{1}{2}.$$
Hope this (little) alternative helps. 
Cheers!
A: Looking at the Taylor's Series Expansion around $a=0$ will give you the answer immediately.  Notice that the taylor series of $\sin(2x)$ is given by
$$\sin(2x)=2x-\frac{4x^3}{3}+\frac{4x^5}{15}-\frac{8x^7}{315}+\dots$$
and for $\cos(3x)$ is
$$\cos(3x)=1-\frac{9x^2}{2}+\frac{27x^4}{8}-\frac{81x^6}{80}+\frac{729x^8}{4480}+\dots$$
Now, multiplying these together gives
$$\sin(2x)\cos(3x)=2x-\frac{18x^3}{2}+\frac{54x^5}{8}+\dots$$
and that's really all we need because there is only one term of this polynomial that has only one $x$ in it (the $2x$).  Now, we divide $x$ by this and take the limit.
$$\lim_{x\to 0}\frac{x}{2x-\frac{18x^3}{2}+\frac{54x^5}{8}+\dots}$$
Multiplying the top and bottom by $\frac{1}{x}, $we have
$$\lim_{x\to 0}\frac{1}{2-\frac{18x^2}{2}+\frac{54x^4}{8}+\dots}=\frac12.$$
A: After l'Hôpital you get $$\lim_{x\to0}\frac{1}{2\cos(2x)\cos(3x)-3\sin(2x)\sin(3x)}=\frac{1}{2-0}=\frac{1}{2}$$.
