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 March 28th, 2017, 10:04 AM #1 Newbie   Joined: Jan 2016 From: england Posts: 17 Thanks: 0 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??
 March 28th, 2017, 10:39 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra 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). Thanks from Skykai
 March 28th, 2017, 10:49 AM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,040 Thanks: 1063 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 Thanks from Skykai

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