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How can I calculate the following limit epsilon-delta definition?

$$\lim_{x \to 0} \left(\frac{\sin(ax)}{x}\right)$$

Edited the equation, sorry...

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    Did you type the problem correctly? Is it supposed to be Sin(ax) / x?2012-11-28
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    I wonder: are you taking $\lim_{y\rightarrow 0}\sin(y)/y=1$ as granted? There is a way to solve this without epsilon and delta, if so...2012-11-28
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    @rschwieb I proved that earlier using epsilon delta, so yes.2012-11-28
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    This kind of questions are odd: if you want an $\,\epsilon-\delta\,$ proof then it is because you already know, or at least heavily suspect, what the limit is...and if you already know/suspect this, it is because you can evaluate the limit by other means, so...2012-11-28
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    @lazyCrab, you proved $\,\frac{\sin y}{y}\xrightarrow [y\to 0]{}1\,$ by epsilon delta?! Would you mind to post your proof, please?2012-11-28
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    @DonAntonio I wouldn't mind doing that, but where should I post it?2012-11-28

2 Answers 2

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you can rewrite $\dfrac{\sin (ax)}{x} = a \dfrac{\sin (ax)}{ax}$, then take limits, as suggested by @rscwieb in the comments

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    Can you please be more specific?2012-11-28
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    so $\lim_{x \rightarrow 0}\dfrac{\sin (ax)}{x} = a \lim_{x \rightarrow 0} \dfrac{\sin (ax)}{ax} = a \cdot 1 = a$? I was hoping you'll get it as it looks like homework2012-11-28
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    Thanks, just wanted to be sure. Though this does not use epsilon delta at all.2012-11-28
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This is easy if you know the power series of the sine function.

$$\sin(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} +\cdots$$

Then $$\sin(ax) = a x - \frac{a^3 x^3}{6} + \frac{a^5 x^5}{120} -\cdots$$

and $\frac{\sin(ax)}{x} = a - \frac{a^3 }{6}x^2 + \frac{a^5 }{120}x^4 -\cdots$ for all $x \neq 0$. Since power series are continuous you can find the limit for $x \rightarrow 0$ by simply setting $x=0$ in this expression, so the limit is $a$.

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    Sorry, haven't "formally" learnt it yet.2012-11-28