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How is the limit of :

$$\lim_{x \rightarrow 0}\frac{\sin(ax)}{\sin(bx)}$$ found without using L'Hopital's rule. I tried substituting $\tan(ax)\cos(ax)$ for $\sin(ax)$, but did not get the answer.

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closed as off-topic by Nosrati, RRL, Alexander Gruber Jan 17 at 23:23

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  • $\begingroup$ your limit is a/b $\endgroup$ – zeraoulia rafik Jul 25 '17 at 23:10
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    $\begingroup$ As is, this question is a problem statement question, which are discouraged. Please edit the question to improve it. $\endgroup$ – Simply Beautiful Art Jul 25 '17 at 23:32
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    $\begingroup$ Seems like this question plays out two groups of users: 'The "help vampires" who flood the site with bad/duplicate questions who only want their question answered and care nothing for the site. The "repwhores" who answer everything they can (or can't).' Don't try shaming me; I didn't write these words: They are part of a post on meta.se. I would personally add a category of spoon-feeders , who insult the asker by assuming they're helpless; pity them $\endgroup$ – Namaste Jul 26 '17 at 0:39
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$$\frac{\sin(ax)}{\sin(bx)}=\frac{\sin(ax)}{ax}\frac{bx}{\sin(bx)}\frac{a}{b}$$

Since $\frac{\sin(ax)}{ax}\to1$ and $\frac{bx}{\sin(bx)}\to1$ our limit tends to $a/b$.

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