Finding trigonometric limit I stuck when finding following limit.
$$\lim_{x\to \pi/4} (\sin 5x-\cos 5x)/(\sin 3x+\cos 3x)$$
 A: Hint: $sin(a)\pm cos(a)$ can be rewriten in the form $A sin(a+da)$.
Then you can study the numerator and denominator left to $\frac \pi 4$
A: Hint. You have
$$
\begin{align}
\cos a-\sin a&=\sqrt{2}\sin\left(\frac{\pi}4-a\right)\\
\cos a+\sin a&=\sqrt{2}\sin\left(\frac{\pi}4+a\right).
\end{align}
$$
A: Let $t=x-\frac{\pi}{4}$.
$$\lim_{x\to \frac{\pi}{4}} \frac{\sin 5x-\cos 5x}{\sin 3x+\cos 3x}=\lim_{t\to 0}\frac{\sin\left(5t+\frac{5\pi}{4}\right)-\cos\left(5t+\frac{5\pi}{4}\right)}{\sin\left(3t+\frac{3\pi}{4}\right)+\cos\left(3t+\frac{3\pi}{4}\right)}$$
$$=\lim_{t\to 0}\frac{-\frac{1}{\sqrt{2}}(\sin (5t)+\cos (5t))-\frac{1}{\sqrt{2}}(\sin (5t)-\cos (5t))}{\frac{1}{\sqrt{2}}(\cos (3t)-\sin (3t))-\frac{1}{\sqrt{2}}(\cos (3t)+ \sin (3t))}$$
$$=\lim_{t\to 0}\frac{\sin (5t)}{\sin (3t)}=\lim_{t\to 0}\left(\frac{\sin(5t)}{5t}\right)\left(\frac{3t}{\sin(3t)}\right)\left(\frac{5}{3}\right)$$
A: HINT:
$$\lim_{x\to\frac{\pi}{4}}\frac{\sin(5x)-\cos(5x)}{\sin(3x)+\cos(3x)}=$$
$$\lim_{x\to\frac{\pi}{4}}\frac{\frac{\text{d}}{\text{d}x}\left(\sin(5x)-\cos(5x)\right)}{\frac{\text{d}}{\text{d}x}\left(\sin(3x)+\cos(3x)\right)}=$$
$$\lim_{x\to\frac{\pi}{4}}\frac{5\cos(5x)+5\sin(5x)}{3\cos(3x)-3\sin(3x)}=$$
$$\lim_{x\to\frac{\pi}{4}}\frac{5(\cos(5x)+\sin(5x))}{3(\cos(3x)-\sin(3x))}$$
