# Find limit using Maclaurin series (remember the importance of big O notation)

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.

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

Using the Maclaurin series of $\tan x$ and letting $x\leftarrow 6x$ yields
• @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 $x \to 0$.) – Meshal Oct 11 '15 at 21:25