I am having difficulties to find the limit for
$$\lim_{x \to 0} \frac{x \cdot \operatorname{cosec}(2x)}{\cos(5x)}$$
I tried to get rid of $ \operatorname{cosec} $ fist
$$\lim_{x \to 0} \frac{\dfrac{x}{\sin(2x)}}{\cos(5x)}$$
Probably I should get it to a point where I could make use of $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$ but I don't know how to continue.
Maybe if one could give me a hint for the next step?
$$\frac{x\csc x}{\cos 2x}=\frac{2x}{\sin 2x}\frac1{2\cos 5x}\xrightarrow[x\to 0]{}1\cdot\frac12$$
$$\lim_{x \to 0} \frac{x \csc(2x)}{\cos(5x)}=\lim_{x \to 0} \frac{x }{\sin2x\cos(5x)}$$ $$=\lim_{x\to0}\frac{2x}{\sin(2x)}\frac{1}{2\cos(5x)}=\frac{1}{2}$$
$$\lim_{x\to 0} \frac{x}{\sin (2x)\cos (5x)}$$ apply the formula $$\sin A\cos B=\frac{1}{2}(\sin (A+B)+\cos (A-B))$$ $$\lim_{x\to 0} \frac{2x}{\sin (7x)-\sin (3x)}$$ further use L hospital to get the limit as $\frac{1}{2}$.
The Taylor expansion of Csc[y] built around y=0 is 1/y + y/6. You may forget the denominator and arrive to x (1/(2x) + (x/3)) = 1/2
