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Is this the correct way to solve for the limit?

I found the limit $x\to 0$ of $\sin4x/ \tan7x$ by these steps

1) $\dfrac{\sin4x}{1} \cdot \dfrac{\cos7x}{\sin7x}$

2) I crossed out the $\sin x$ in the numerator and denominator leaving me with $\dfrac{4\cos7x}{7}$

3) $4 (\cos7(0)) = 4 (\cos0=1)$

4) I was left with $ 4/7$

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    This would be a good response for [this](http://math.stackexchange.com/questions/260656/cant-argue-with-success-looking-for-bad-math-that-gets-away-with-it) post.2012-12-18
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    No, your reasoning to get to 2) is not correct. Given $\sin (4x)\over \sin(7x)$ you cannot "cross out the $\sin x$" to obtain ${\sin (4x)\over \sin(7x)}={4\over 7}$. This is not true...2012-12-18
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    If this is not the correct method, how would you solve this example @David Mitra2012-12-18
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    As in the answers below. What would be correct is to say is that $\lim_{x\rightarrow0}{\sin(4x)\over\sin(7x)}={4\over7}$. To prove this, you'd use the fact (which you should have encountered if this problem is posed to you) that $\lim_{x\rightarrow0}{\sin(x)\over x}={1}$:$$\lim_{x\rightarrow0}{\sin(4x)\over\sin(7x)}=\lim_{x\rightarrow0}\Bigl[4\cdot{\sin(4x)\over 4x}\cdot {1\over7}\cdot{ 7x\over \cdot\sin( 7x)}\Bigr]=4\cdot 1\cdot{1\over7}\cdot1={4\over7}.$$ Then you can use this to find your limit.2012-12-18
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    @David Mitra Where are you getting the x in the denominator (4x) and numerator (7x) from?2012-12-18
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    Those guys and the constant factors $4$ and ${1\over7}$ *do* cancel (since $x\ne0$); so it's ok to just put them there. They were put there because, then, one can compute the limits $\lim_{x\rightarrow0}{\sin(4x)\over 4x}$ and $\lim_{x\rightarrow0}{7x\over \sin(7x)}$.2012-12-18
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    @user44816 Basically, what is being done is that we *want* to get something of the form $\frac{\sin y}{y}$. Since we have $\sin 4x$, let $y = 4x$. Then $\frac{\sin y}{y} = \frac{\sin 4x}{4x}$. Finally, since $y \to 0$ as $x \to 0$, we don't have to change the target of our limit. Now we have a nice situation where we can use $\lim_{y \to 0} \frac{\sin y}{y} = 1$.2012-12-18

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