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I've been trying to solve this question for hours. It asks to find the limit without L'Hôpital's rule. $$\lim_{x\to0}\frac{1-\cos3x+\sin3x}x$$ Any tips or help would be much appreciated.

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  • $\begingroup$ Are you familiar with derivatives yet? $\endgroup$ – Cameron Williams Oct 7 '16 at 3:03
  • $\begingroup$ Can you use the Taylor expansions of $\cos, \sin$? $\endgroup$ – copper.hat Oct 7 '16 at 3:13
  • $\begingroup$ Comment: You shouldn't use LHR as it would be circular here, do you see why? $\endgroup$ – GFauxPas Oct 7 '16 at 3:24
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    $\begingroup$ @GFauxPas What proof of LHR do you have in mind which would make using it here circular? $\endgroup$ – dxiv Oct 7 '16 at 3:29
  • $\begingroup$ What's with the italics? $\endgroup$ – copper.hat Oct 7 '16 at 3:29
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If you are given that $\lim_{x \to 0}{\sin x \over x } = 1$, then since $1-\cos (3x) = 2 \sin^2 ({3 \over 2} x)$ (half angle formula), we have

\begin{eqnarray} {1 -\cos (3x) +\sin (3x) \over x } &=& {2 \sin^2 ({3 \over 2} x) \over x} +{\sin (3x) \over 3x} {3x \over x} \\ &=& 2 ({\sin ({3 \over 2} x) \over {3 \over 2} x})^2 { ({3 \over 2} x)^2 \over x} + 3 {\sin (3x) \over 3x} \\ &=& {9 \over 2} x ({\sin ({3 \over 2} x) \over {3 \over 2} x})^2 + 3 {\sin (3x) \over 3x} \end{eqnarray} Taking limits gives $3$.

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Do you know how to do without L'Hôpital's rule $$ \lim_{x\to 0}\frac{\sin x}{x} \quad \text{ and }\quad \lim_{x\to 0}\frac{\cos x-1}{x}? $$ Then try the "simpler" version of your problem: $$ \lim_{x\to 0}\frac{\sin 3x}{x}, \qquad \lim_{x\to 0}\frac{\cos 3x-1}{x}. $$

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Since $\lim_{x\to 0}\frac{\sin x}{x}=1$ it follows that $\lim_{x\to 0}\frac{\sin(3x)}{x}=3$. On the other hand, since $$ 1-\cos(3x) = 2\sin^2\left(\frac{3x}{2}\right) $$ it follows that $\lim_{x\to 0}\frac{1-\cos(3x)}{x}=0$. It should not be difficult to finish from here.

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  • $\begingroup$ As an alternative to the double angle formula, one could multiply the numerator and denominator by of the cosine limit by $1+\cos(3x)$ because then you can use that $1-\cos^2(3x)=\sin^2(3x)$. $\endgroup$ – Michael Burr Dec 22 '16 at 18:16
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If I write it in this form, does it look familiar ?

$$\lim_{x\to 0} \frac{[-\cos(3x) + \sin(3x)] - (-1)}{x - 0}$$

$$\lim_{x\to 0} \frac{[-\cos(3x) + \sin(3x)] - [-cos(0) + sin(0)]}{x - 0}$$

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Taylor expansion is always a good solution since the method will provide the limit and more.

Remembering that $$\cos(t)=1-\frac{t^2}{2}+O\left(t^4\right)$$ $$\sin(t)=t-\frac{t^3}{6}+O\left(t^4\right)$$ replace $t$ by $3x$ to get $$1-\cos (3 x)+\sin (3 x)=1-\left(1-\frac{9 x^2}{2}+O\left(x^4\right) \right) +\left( 3 x-\frac{9 x^3}{2}+O\left(x^4\right)\right)$$ $$1-\cos (3 x)+\sin (3 x)=3 x+\frac{9 x^2}{2}-\frac{9 x^3}{2}+O\left(x^4\right)=3 x+\frac{9 x^2}{2}+O\left(x^3\right)$$ $$\frac{1-\cos (3 x)+\sin (3 x) }x=3+\frac{9 x}{2}+O\left(x^2\right)$$ which shows the limit and how it is approached.

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