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Find

$$\lim_{x \to 0}\frac{x-\sin(x)\cos(x)}{\sin(x)-\sin(x)\cos(x)}\;.$$

Applying L'Hopital's rule directly does not seem to get me anywhere. I also tried dividing the numerator and denominator by $\sin(x)$, which did not seem to work.

Is there a some sort of trick I am missing here?

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    $\begingroup$ Recall that $\sin(2x)=2\sin x\cos x$, so $\sin x\cos x=\frac12\sin(2x)$ and consequently $\dfrac{x-\sin x\cos x}{\sin x-\sin x\cos x} = \dfrac{2x - \sin(2x)}{2\sin x-\sin(2x)}$. Then you have to apply L'Hopital's rule three times. $\endgroup$ – Michael Hardy Jan 7 '15 at 21:43
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Instead of using l'Hôpital's rule (which does work here if used repeatedly), the Taylor expansion of the sine function is particularly useful here. So write $$\sin(x)=x-\frac{1}{6}x^3+\mathcal{O}(x^5).$$ Noting that $\sin(x)\cos(x)=\frac{1}{2}\sin(2x)$, we see that $\sin(x)\cos(x)=x-\frac{2}{3}x^3+\mathcal{O}(x^5)$ (by replacing $x$ with $2x$ in the Taylor expansion and then dividing by $2$).
Now we can compute the limit: $$\begin{aligned} \lim_{x\to 0}\frac{x-\sin(x)\cos(x)}{\sin(x)-\sin(x)\cos(x)}&=\lim_{x\to 0}\frac{x-(x-\frac{2}{3}x^3+\mathcal{O}(x^5))}{x-\frac{1}{6}x^3+\mathcal{O}(x^5)-(x-\frac{2}{3}x^3+\mathcal{O}(x^5))}\\ &=\lim_{x\to 0}\frac{\frac{2}{3}x^3+\mathcal{O}(x^5)}{\frac{1}{2}x^3+\mathcal{O}(x^5)}\\ &=\lim_{x\to 0}\frac{\frac{2}{3}+\mathcal{O}(x^2)}{\frac{1}{2}+\mathcal{O}(x^2)}=\frac{4}{3}. \end{aligned}$$

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  • $\begingroup$ what you probably meant was $\sin x = x -\frac{x^3}{3!} + O(x^5)$ $\endgroup$ – Alex Jan 7 '15 at 22:18
  • $\begingroup$ No, I did not mean that; there is no need for using 5th order terms, is there? $\endgroup$ – Joffysloffy Jan 7 '15 at 22:22
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    $\begingroup$ I agree. All I meant that $\frac{O(x^3)}{6}$ is a strange notation. Usually we either have $\frac{x^3}{3!}$ or $O(x^3)$ since $O(\cdot)$ implies an asymptotic constant $\endgroup$ – Alex Jan 7 '15 at 22:27
  • $\begingroup$ Oh, I see. Yes, now that you mentioned it, it looks a little odd. I'll change that, thanks! $\endgroup$ – Joffysloffy Jan 7 '15 at 22:28
  • $\begingroup$ you are welcome $\endgroup$ – Alex Jan 7 '15 at 22:30
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$$\lim_{x \to 0}\frac{x-\sin(x)\cos(x)}{\sin(x)-\sin(x)\cos(x)}=\lim_{x \to 0}\frac{x-\frac{\sin(2x)}{2}}{\sin(x)-\frac{\sin(2x)}{2}}$$

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  • $\begingroup$ this isn't the final answer is it? $\endgroup$ – Irrational Person Jan 7 '15 at 21:56
  • $\begingroup$ @Bot no, then you have to apply L'Hopital but is too boring :P $\endgroup$ – rlartiga Jan 7 '15 at 22:01
  • $\begingroup$ i agree with you on that. $\endgroup$ – Irrational Person Jan 7 '15 at 22:01
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1) Apply L'Hopital's rule

2) $\cos^2(x)-\sin^2(x)=\cos(2x)$

3) Apply L'Hopital's rule (2nd time)

4) Apply L'Hopital's rule (3rd time)

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You can use the known facts that $$ \lim_{x\to0}\frac{\sin x}{x}=1 \qquad\text{and}\qquad \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2} $$ or $$ \lim_{x\to0}\frac{x}{\sin x}=1 \qquad\text{and}\qquad \lim_{x\to0}\frac{x^2}{1-\cos x}=2 $$

You can then write $$ \lim_{x \to 0}\frac{x-\sin(x)\cos(x)}{\sin(x)-\sin(x)\cos(x)}= \lim_{x\to0} \frac{x-\sin(x)\cos(x)}{x^3} \frac{x}{\sin x} \frac{x^2}{1-\cos x} $$ so you just need to compute $$ 2\lim_{x\to0}\frac{x-\sin(x)\cos(x)}{x^3} $$

Now \begin{align} 2\lim_{x\to0}\frac{x-\sin(x)\cos(x)}{x^3} &=2\lim_{x\to0}4\frac{2x-2\sin x\cos x}{8x^3}&&\text{(set $2x=t$)}\\[2ex] &=8\lim_{t\to0}\frac{t-\sin t}{t^3}&&\text{(apply l'Hôpital)}\\[2ex] &=8\lim_{t\to0}\frac{1-\cos t}{3t^2}&&\text{(known limit)}\\[2ex] &=8\cdot \frac{1}{3}\cdot\frac{1}{2}=\frac{4}{3} \end{align}

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Apply L'Hospital to equate the red parts: $$ \begin{align} \lim_{x\to0}\frac{x-\sin(x)\cos(x)}{\sin(x)-\sin(x)\cos(x)} &=\lim_{x\to0}\left(1+\frac{x-\sin(x)}{\sin(x)(1-\cos(x))}\right)\\ &=\lim_{x\to0}\left(1+\color{#C00000}{\frac{x-\sin(x)}{\sin^3(x)}}(1+\cos(x))\right)\\ &=\lim_{x\to0}\left(1+\color{#C00000}{\frac{1-\cos(x)}{3\sin^2(x)}}\frac{1+\cos(x)}{\color{#C00000}{\cos(x)}}\right)\\ &=\lim_{x\to0}\left(1+\frac1{3\cos(x)}\right)\\[3pt] &=\frac43 \end{align} $$

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I like to do these by expanding power series about $x=0$. It helps to notice that $\sin x \cos x = \frac{1}{2} \sin (2x)$.

But if you L'Hospital 3 times you get $$ \frac{4 \cos 2x}{4 \cos 2x - \cos x} \rightarrow \frac{4}{3}$$

That is, $$ \frac{x-\frac{1}{2}\sin(2x)}{\sin x-\frac{1}{2}\sin(2x)} \\ \frac{1-\cos(2x)}{\cos x-\cos(2x)} \\ \frac{2\sin(2x)}{2\sin(2x)-\sin x}\\ \frac{4\cos(2x)}{4\cos(2x)-\cos x} \rightarrow \frac{4}{3} $$

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$$\frac{x-\sin(x)\cos(x)}{\sin(x)-\sin(x)\cos(x)}=\frac{x-\sin(x)}{\sin(x)-\sin(x)\cos(x)}+1=\frac{x-\sin(x)}{\sin(x)2\sin^2(x/2)}+1.$$

Then, $\sin x=x+o(x^3)=x-x^3/6+o(x^5)$ and the limit is that of $$\frac{\frac{x^3}6}{\frac{x^3}2}+1.$$

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Just as Joffysloffy answered, Taylor series are extremely convenient for solving the problem of the limit and even more.

Rewrite, as Michael Hardy suggested, $$A=\dfrac{x-\sin x\cos x}{\sin x-\sin x\cos x} = \dfrac{2x - \sin(2x)}{2\sin x-\sin(2x)}$$ and use the fact that $$\sin(y)=\sum_{i=0}^{\infty}\frac{(-1)^n}{(2n+1)!}y^{2n+1}$$ So, the numerator write $$\frac{4 x^3}{3}-\frac{4 x^5}{15}+\frac{8 x^7}{315}+O\left(x^8\right)$$ and the denominator $$x^3-\frac{x^5}{4}+\frac{x^7}{40}+O\left(x^8\right)$$ So, performing the long division $$A=\frac{4}{3}+\frac{x^2}{15}+\frac{11 x^4}{1260}+O\left(x^5\right)$$ which gives not only the limit but also how it is approached.

If you plot the function and the approximation for $0 \leq x \leq \frac{\pi}{2}$, you will probably be amazed to notice how close are the two curves.

What is interesting (at least to me !) is that, if you had to solve for $x$ the equation $A=\frac{3}{2}$, the approximation limits the problem to a quadratic in $x^2$ and the solution would be $$x=\sqrt{\frac{1}{11} \left(\sqrt{4074}-42\right)}\approx 1.40867$$ while the exact solution is $\approx 1.37873$. Using only the second order approximation would immediately lead to $x=\sqrt{\frac{5}{2}}\approx 1.58114$ which already a good approximation from which any root-finding method could efficiently start.

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