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I have a problem that sounds like this:

Find the limit $$\lim_{x\rightarrow 0} \frac{14\tan(6x)-84x}{6x^3}$$ using Maclaurin series, and don't forget the importance of big O notation.

I have tried to find the Maclaurin series in different ways, but I always end up with the wrong answer. And I don't know how to use the big O notation in a helpful way here.

Thank you in advance.

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    $\begingroup$ As is, you have a problem, the limit doesn't exist. I'm pretty sure it should be $$\frac{14\tan (6x) - 84x}{6x^3}.$$ $\endgroup$ – Daniel Fischer Oct 10 '15 at 21:56
  • $\begingroup$ It should be $84x$ rather than $84$. $\endgroup$ – Meshal Oct 10 '15 at 22:02
  • $\begingroup$ You're right, I wrote it wrong. $\endgroup$ – netwon1227 Oct 10 '15 at 22:33
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Using the Maclaurin series of $\tan x$ and letting $x\leftarrow 6x$ yields

\begin{align} \lim_{x\rightarrow 0} \frac{14\tan(6x)-84x}{6x^3}&=\lim_{x\rightarrow 0} \frac{14\left \{ 6x+72x^3+O(x^5) \right \}-84x}{6x^3}\\ &=\lim_{x\rightarrow 0} \frac{84x+1008x^3+O(x^5)-84x}{6x^3}\\ &=168 \end{align}

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  • $\begingroup$ Thanks, after trying to calculate it for like one hour, it's nice to see how easy it really is... $\endgroup$ – netwon1227 Oct 10 '15 at 22:51
  • $\begingroup$ @netwon1227 You are welcome. $\endgroup$ – Meshal Oct 10 '15 at 22:53
  • $\begingroup$ Just to be sure, you ignore O(x^5) because it is very small? $\endgroup$ – netwon1227 Oct 11 '15 at 21:09
  • $\begingroup$ @netwon1227 No. Because if you work out the second step in details, you will end up with $\frac{1008x^3}{6x^3} + \frac{O(x^5)}{6x^3} = 168 + O(x^2)$. Now taking the limit $x \to 0$ of this will yield $168$ (since any term of the form $O(x^p)$ with $p \ge 1$ will give zero as $x \to 0$.) $\endgroup$ – Meshal Oct 11 '15 at 21:25

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