Calculate $\lim_{h\to 0} \frac{\cos(x-2h)-\cos(x+h)}{\sin(x+3h)-\sin(x-h)}$ Calculate $$\lim_{h\to 0} \frac{\cos(x-2h)-\cos(x+h)}{\sin(x+3h)-\sin(x-h)}$$
If I take the limit it results in undefined value.
I try to change the formation using identity $\sin A + \sin B$
$$\lim_{h\to 0} \frac{-2\sin\frac{(2x-h)}2\sin(-3h/2)}{2\cos(x+h)\sin(2h)}$$
How do I actually evaluate the limit? With and without derivative?
 A: Now, $$\frac{-2\sin\frac{2x-h}2\sin(-\frac{3h}{2})}{2\cos(x+h)\sin2h}= \frac{-2\sin\frac{(2x-h)}2\frac{\sin(-\frac{-3h}{2})}{-\frac{3h}{2}}}{2\cos(x+h)\frac{\sin2h}{2h}}\cdot\frac{-\frac{3}{2}}{2}\rightarrow\frac{3\sin{x}}{4\cos{x}}$$
A: You need to use $$\lim_{y\to 0}\frac{\sin y}y=1$$
Then $$\lim_{h\to 0}\frac{\sin(-3h/2)}{\sin(2h)}=\lim_{h\to 0}\frac{\sin(-3h/2)}{-3h/2}\frac{2h}{\sin(2h)}\frac{-3h/2}{2h}=-\frac 34$$
A: Using Taylor series
$$A=\frac{\cos(x-2h)-\cos(x+h)}{\sin(x+3h)-\sin(x-h)}$$
$$\cos(x-2h)=\cos (x)+2 h \sin (x)-2 h^2 \cos (x)+O\left(h^3\right)$$
$$\cos(x+h)=\cos (x)-h \sin (x)-\frac{1}{2} h^2 \cos (x)+O\left(h^3\right)$$
$$\sin(x+3h)=\sin (x)+3 h \cos (x)-\frac{9}{2} h^2 \sin (x)+O\left(h^3\right)$$
$$\sin(x-h)=\sin (x)-h \cos (x)-\frac{1}{2} h^2 \sin (x)+O\left(h^3\right)$$
$$A=\frac{3}{4} \tan (x)+\frac{1}{8}  \left(6 \tan ^2(x)-3\right)h+O\left(h^2\right)$$ shows the limit and how it is approached.
A: Direct evaluation gives $0/0$ so apply L'Hospital's rule:
\begin{align}\lim_{h\to 0}\frac{\frac{d}{dh}\left(\cos(x-2h)-\cos(x+h)\right)}{\frac{d}{dh}\left(\sin(x+3h)-\sin(x-h)\right)}&=\lim_{h\to 0}\frac{2\sin(x-2h)+\sin(x+h)}{3\cos(x+3h)+\cos(x-h)}\\&=\frac{2\sin(x)+\sin(x)}{3\cos(x)+\cos(x)}\\&=\frac{3}{4}\tan(x)
\end{align}
