# Limit of a quotient without using L'Hopital rule [closed]

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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.

• your limit is a/b – zeraoulia rafik Jul 25 '17 at 23:10
• As is, this question is a problem statement question, which are discouraged. Please edit the question to improve it. – Simply Beautiful Art Jul 25 '17 at 23:32
• 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 – Namaste Jul 26 '17 at 0:39

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