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March 28th, 2017, 10:04 AM   #1
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Maclaurin expansions, trig and limits all in one!

Find $\displaystyle \displaystyle \lim_{x\rightarrow 0}\frac{1-cos(4x)+xsin(3x)}{x^2}$

Any suggestions on how I can answer this question using the Maclaurin expansion please?

I started differentiating and got:

$\displaystyle f'x) = -2x^-3 (1-cos(4x) + xsin(3x))+x^-2 (3xcos(3x)+4sin(4x)+sin(3x))$

and also f''(x) but if I sub in 0 it will always give me 0 so I must be doing something wrong surely??
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March 28th, 2017, 10:39 AM   #2
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Put in the first few terms of the Maclaurin series for $\cos{(4x)}$ and the first few terms of the Maclaurin series for $\sin{(3x)}$ and then simplify. (Two terms each should be sufficient).
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March 28th, 2017, 10:49 AM   #3
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if you are going to mess w/series expansions it will behoove you to learn these two formulae

$\cos(x) = \displaystyle \sum_{k=0}^\infty~\dfrac{x^{2k}}{(2k)!}$

$\sin(x) = \displaystyle \sum_{k=0}^\infty~\dfrac{x^{2k+1}}{(2k+1)!}$

your expression then becomes

$\displaystyle \lim_{x\to 0} \dfrac{1-\displaystyle \sum_{k=0}^\infty~\dfrac{(4x)^{2k}}{(2k)!}+x \displaystyle \sum_{k=0}^\infty~\dfrac{(3x)^{2k+1}}{(2k+1)!}}{x^ 2}=$

and I'll let you mess with simplifying all this.

what you are looking for is the remaining constant terms as all the terms with powers of $x$ will tend to 0
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