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I stuck when finding following limit.

$$\lim_{x\to \pi/4} (\sin 5x-\cos 5x)/(\sin 3x+\cos 3x)$$

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  • $\begingroup$ What are your thoughts? Which methods are allowable? In particular, can we use L'Hopital's rule? $\endgroup$ Nov 8, 2015 at 0:40

4 Answers 4

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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 $\frac \pi 4$

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Hint. You have $$ \begin{align} \cos a-\sin a&=\sqrt{2}\sin\left(\frac{\pi}4-a\right)\\ \cos a+\sin a&=\sqrt{2}\sin\left(\frac{\pi}4+a\right). \end{align} $$

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Let $t=x-\frac{\pi}{4}$.

$$\lim_{x\to \frac{\pi}{4}} \frac{\sin 5x-\cos 5x}{\sin 3x+\cos 3x}=\lim_{t\to 0}\frac{\sin\left(5t+\frac{5\pi}{4}\right)-\cos\left(5t+\frac{5\pi}{4}\right)}{\sin\left(3t+\frac{3\pi}{4}\right)+\cos\left(3t+\frac{3\pi}{4}\right)}$$

$$=\lim_{t\to 0}\frac{-\frac{1}{\sqrt{2}}(\sin (5t)+\cos (5t))-\frac{1}{\sqrt{2}}(\sin (5t)-\cos (5t))}{\frac{1}{\sqrt{2}}(\cos (3t)-\sin (3t))-\frac{1}{\sqrt{2}}(\cos (3t)+ \sin (3t))}$$

$$=\lim_{t\to 0}\frac{\sin (5t)}{\sin (3t)}=\lim_{t\to 0}\left(\frac{\sin(5t)}{5t}\right)\left(\frac{3t}{\sin(3t)}\right)\left(\frac{5}{3}\right)$$

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HINT:

$$\lim_{x\to\frac{\pi}{4}}\frac{\sin(5x)-\cos(5x)}{\sin(3x)+\cos(3x)}=$$ $$\lim_{x\to\frac{\pi}{4}}\frac{\frac{\text{d}}{\text{d}x}\left(\sin(5x)-\cos(5x)\right)}{\frac{\text{d}}{\text{d}x}\left(\sin(3x)+\cos(3x)\right)}=$$ $$\lim_{x\to\frac{\pi}{4}}\frac{5\cos(5x)+5\sin(5x)}{3\cos(3x)-3\sin(3x)}=$$ $$\lim_{x\to\frac{\pi}{4}}\frac{5(\cos(5x)+\sin(5x))}{3(\cos(3x)-\sin(3x))}$$

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