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I wonder how to do this in different way from L'Hôpital's rule:

$$\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}.$$

Please help me solve this without using L'Hopital's rule.

Mang Sopheak
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5 Answers5

6

Instead of using power series, which feels like a way of hiding the use of derivatives and, ultimately, L'Hôpital, we can use a trigonometric identity and take limits.

Since $$ \frac{4\sin(x)-2\sin(2x)}{2\sin(2x)-\sin(4x)} =\frac1{2(1+\cos(x))\cos(x)}\tag{1} $$ we have $$ \lim_{x\to0}\frac{2^{k+2}\sin\left(\frac{x}{2^{k+2}}\right)-2^{k+1}\sin\left(\frac{x}{2^{k+1}}\right)}{2^{k+1}\sin\left(\frac{x}{2^{k+1}}\right)-2^k\sin\left(\frac{x}{2^k}\right)} =\frac14\tag{2} $$ Taking the product of $(2)$ from $k=0$ to $k=n-1$, we get $$ \lim_{x\to0}\frac{2^{n+1}\sin\left(\frac{x}{2^{n+1}}\right)-2^n\sin\left(\frac{x}{2^n}\right)}{2\sin\left(\frac x2\right)-\sin\left(x\right)} =\frac1{4^n}\tag{3} $$ Summing $(3)$ from $n=0$ to $n=\infty$, we get $$ \lim_{x\to0}\frac{x-\sin(x)}{2\sin\left(\frac x2\right)-\sin\left(x\right)}=\frac43\tag{4} $$ Applying $(1)$ gives $$ \begin{align} \lim_{x\to0}\frac{2\sin\left(\frac x2\right)-\sin\left(x\right)}{2\sin(x)-\sin(2x)} &=\lim_{x\to0}\frac12\frac{4\sin\left(\frac x2\right)-2\sin\left(x\right)}{2\sin(x)-\sin(2x)}\\ &=\frac18\tag{5} \end{align} $$ Multiplying $(4)$ and $(5)$, and taking the reciprocal, yields $$ \begin{align} \lim_{x\to0}\frac{2\sin(x)-\sin(2x)}{x-\sin(x)}=6\tag{6} \end{align} $$

robjohn
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  • What inspired this development? It is quite an interesting approach, but requires more than a bit of inspiration (and/or hindsight). It certainly is not the most efficient way of tackling this without "The Hospital." But I just have to give it a +1 for originality! – Mark Viola Jul 09 '15 at 05:25
  • @Dr.MV: it depends on what you intend by "without L'Hôpital". The usual reason I see for avoiding L'Hôpital is that we are dealing with pre-calculus, and power series are as foreign as L'Hôpital. The tools used above, trigonometric identities, geometric sums, and telescoping sums, are accessible at a pre-calculus level. – robjohn Jul 09 '15 at 09:01
  • Yes. But what inspired you here. That is one very slick development. Nothing hard to explain, but what would motivate you to initiate? It's really neat. – Mark Viola Jul 09 '15 at 12:37
  • It is easy to get $\lim\limits_{x\to0}\frac{\sin(x)}x$ geometrically, but $\frac{x-\sin(x)}{\sin^3(x)}$ is harder to control. Using $x=\lim\limits_{n\to\infty}2^n\sin(x2^{-n})$ seemed a way to go. I also used a similar idea in this answer. – robjohn Jul 09 '15 at 13:34
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Write \begin{eqnarray*} \frac{2\sin x-\sin 2x}{x-\sin x} &=&\frac{2\sin x-2\sin x\cos x}{x-\sin x} \\ &=&2\left( \frac{\sin x}{x}\right) \left( \frac{1-\cos x}{x^{2}}\right) \left( \frac{x^{3}}{x-\sin x}\right) \end{eqnarray*} Using standard limits \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1 \\ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2} \\ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}} &=&\frac{1}{6} \end{eqnarray*} it follows that \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{2\sin x-\sin 2x}{x-\sin x} &=&2\lim_{x\rightarrow 0}\left( \frac{\sin x}{x}\right) \lim_{x\rightarrow 0}\left( \frac{1-\cos x}{% x^{2}}\right) \lim_{x\rightarrow 0}\left( \frac{x^{3}}{x-\sin x}\right) \\ &=&2\left( 1\right) \left( \frac{1}{2}\right) \left( \frac{6}{1}\right) \\ &=&6. \end{eqnarray*}

Idris Addou
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2

We have

$$\sin x=x-\frac16x^3+O(x^5)\implies 2\sin x-\sin 2x =x^3+O(x^5)$$

and

$$x-\sin x=\frac16x^3+O(x^5)$$

Thus

$$\frac{2\sin x-\sin 2x}{x-\sin x}=6+O(x^2)\to 6\,\,\text{as}\,\,x\to 0$$

Mark Viola
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2

Just as Corindo and Dr. MV answered, for this kind of problems, Taylor expansions are very useful not only for obtaing the limit but also to detect how it is approached.

Using Taylor series for $$\sin(y)=y-\frac{y^3}{6}+\frac{y^5}{120}+O\left(y^6\right)$$ and replacing $y$ by $2x$ in numerator and $x$ in denominator, we then have $$A=\frac{2\sin x-\sin 2x}{x-\sin x}=\frac{2(x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots)-(2 x-\frac{4 x^3}{3}+\frac{4 x^5}{15}+\cdots)}{x-(x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots)}$$ $$A=\frac{x^3-\frac{x^5}{4}+\cdots}{\frac{x^3}{6}-\frac{x^5}{120}+\cdots}$$ Performing the long division, we then have $$A=6-\frac{6 x^2}{5}+\cdots$$

If you plot on the same graph the original function and this last approximation for the range $-1\leq x \leq 1$, you will probably be amazed to see how close the two curves are.

Claude Leibovici
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    I wonder if there is a way to do it with just trig identities and the basic trig limits $\frac{\sin x}{x}\to 1$ and $\frac{1-\cos x}{x}\to 0$, avoiding Taylor expansions. – G Tony Jacobs Jul 07 '15 at 03:10
  • I totally agree with you as long as we just speak about the limit. What I wanted to show is that we can get much more information using Taylor. – Claude Leibovici Jul 07 '15 at 03:13
  • Yes, the Taylor series is extremely useful. I've been playing with this limit, and managed to show that $\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}$ is equal to $\frac14\lim_{x\to 0}\frac{3\sin x-\sin3x}{x-\sin x}$. This seems to be a step in the wrong direction, but possibly an interesting one. – G Tony Jacobs Jul 07 '15 at 03:16
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Use this 2 relations: $$\sin(x) = x - \frac{x^3}{6} + o(x^4) = x + o(x^2)$$ $$ \cos(x) = 1 - \frac{x^2}{2} + o(x^3)$$

corindo
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