Given the $\lim_{x\to \pi} \dfrac{\sin(3x)}{\sin(5x)}$. The answer is $\dfrac 3 5$ but what are the solution steps?
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3$\begingroup$ Write $x = \pi+h$, then you get $\lim\limits_{h\to 0} \frac{\sin (3(\pi+h))}{\sin (5(\pi+h))}$. Using $\sin (\pi+\xi) = -\sin\xi$ then gets you a simpler form. $\endgroup$– Daniel FischerNov 17, 2013 at 16:50
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$\begingroup$ Do you want to know how to prove $\lim \limits_{x\to \pi}\left(\dfrac{\sin(3x)}{\sin(5x)}\right)=\dfrac 3 5?$ $\endgroup$– Git GudNov 17, 2013 at 16:50
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2$\begingroup$ Taylor Series? L'Hopital? First principles? We don't know what tools you have knowledge of. $\endgroup$– preferred_anonNov 17, 2013 at 16:53
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$\begingroup$ Simpler - better. $\endgroup$– J.OlufsenNov 17, 2013 at 16:56
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4 Answers
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Use L'Hopital's rule (L.H.):
$$\lim_{x\to \pi} \dfrac{\sin(3x)}{\sin(5x)} \quad \overset{L.H.}{=} \quad \lim_{x \to \pi}\dfrac{\Big(\sin(3x)\Big)'}{\Big(\sin(5x)\Big)'} = \lim_{x\to \pi} \dfrac {3\cos 3x}{5 \cos 5x} = \dfrac {3\cdot 1}{5 \cdot 1} = \dfrac 35$$
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Hint: Use $$\lim_{x\rightarrow 0} \frac {sinx}{x}=1$$ and use sin($\pi$-x)=sinx
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Put $y=x-\pi$ and observing that $\sin(\theta+n\pi)=(-1)^n\sin \theta$, you have $$ \lim_{x\to\pi}\frac{\sin(3x)}{\sin(5x)}=\lim_{y\to 0}\frac{(-1)^3\sin(3y)}{(-1)^5\sin(5y)}=\lim_{y\to 0}\frac{3y\frac{\sin(3y)}{3y}}{5y\frac{\sin(5y)}{5y}}=\lim_{y\to 0}\frac{3}{5}\frac{\frac{\sin(3y)}{3y}}{\frac{\sin(5y)}{5y}}=\frac{3}{5} $$ using the fundamental limit $\displaystyle\lim_{z\to 0}\tfrac{\sin(z)}{z}=1$.
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We know that while $x\sim 0$ then $\sin(ax)\sim(ax)$. See An Infinitesimal Function or Classification of Infinitesimal Functions for more details.
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$\begingroup$ @RCola: Yes and that is why you should take Daniel's leading hint. :) $\endgroup$– MikasaNov 17, 2013 at 17:15