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I am new to this site, so I don't know if this will appear correctly.

I need to solve this limit without L'Hospital's Rule:

$$ \lim_{x\to 0}\frac{x-\sin x}{x^3}.$$

I know that the result is $\frac{1}{6}$ but I need a step-by-step solution.

Thanks in advance.

Jack D'Aurizio
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Mohammed Masri
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  • oh ok thanks for the tip :) – Mohammed Masri Oct 12 '15 at 17:24
  • http://math.stackexchange.com/questions/134051/solving-lim-limits-x-to0-fracx-sinxx2-without-lhospitals-rule?lq=1 ...correction. There is actually an answer for this here. – randomgirl Oct 12 '15 at 17:44
  • I found the link to the above question here: https://mathindex.wordpress.com/ You may want to bookmark this site if you are going to be looking for other limits. –  Oct 12 '15 at 17:52
  • Simplify the entire fraction by x, then use the fact that $\dfrac{\sin x}x\simeq\cos x$ around the origin $($plot the graphics of the two functions to visualize this$)$. We also have $1-\cos2t=2\sin^2t$, and $\lim\limits_{t\to0}\dfrac{\sin t}t=1$. – Lucian Oct 13 '15 at 02:10

1 Answers1

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HINT Expand $\sin x$ into Taylor series around 0

gt6989b
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  • Isn't that just l'Hôpital's rule in disguise? I'm not saying there's a better way, though. –  Oct 12 '15 at 17:27
  • I am sorry but I have no idea what Taylor series is .. I am a medical student and I need to solve this as a part of a quiz for a friend and he is in high school, I asked him about taylor series, he doesn't know about that as well.. I am sorry but is there a differnet way? thank you so much – Mohammed Masri Oct 12 '15 at 17:32
  • @MikeHaskel interesting thought... series works sometimes when Taylor's theorem cannot be applied, it is more generic I think. But is very similar in style... – gt6989b Oct 12 '15 at 18:01
  • @MikeHaskel there is a different way using algebraic manipulation: http://math.stackexchange.com/a/217086/16192 – gt6989b Oct 12 '15 at 18:03