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Can you please help me solve:

$$\lim_{x \rightarrow 0} \frac{1- \cos x}{x \sin x}$$

Every time I try to calculate it I find another solution and before I get used to bad habits, I'd like to see how it can be solved right, so I'll know how to approach trigonometric limits.

I tried to convert $\cos x$ to $\sin x$ by $\pi -x$, but I think it's wrong. Should I use another identity?

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  • $\begingroup$ For future reference, here is a page where you can read up on how to format math on this site. $\endgroup$ – Arthur May 23 '17 at 15:06
  • $\begingroup$ Solving it in as many ways as you can is a GOOD thing. There is no surefire approach to limits. Nevertheless, assuming you have shown that $\lim_{x \to 0} \frac{\sin(x)}{x}=1$ already then you can use LHopital here, which is a generally good way to approach these. Even better, you could use series expansions, which solve this trivially $\endgroup$ – Brevan Ellefsen May 23 '17 at 15:08
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Whilst the other answers are 'clever', note that these sort of limits can be done automatically $$\frac{1-\cos x}{x\sin x} = \frac{x^2/2 + O(x^4)}{x^2 + O(x^4)} = \frac{1/2 + O(x^2)}{1 + O(x^2)} \stackrel{x \to 0}{\longrightarrow} \frac{1}{2}$$

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  • $\begingroup$ Well you're in the realm of Taylor with that, which is not needed, because this can be done in the same week you treat derivatives of trig functions. $\endgroup$ – zhw. May 23 '17 at 19:40
  • $\begingroup$ @zhw. or you could define $\sin$ and $\cos$ as power series and avoid all the potentially circular proofs of $\sin x/x \to 0$. $\endgroup$ – Zain Patel May 23 '17 at 19:42
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    $\begingroup$ @zhw: We learn all of this machinery because it makes it easier to solve problems! There is no point in doing things the hard way simply because you can. $\endgroup$ – Hurkyl Jul 11 '17 at 5:47
  • $\begingroup$ @Hurkyl i don't think Taylor makes it easier. All you need is $(\sin x)/x\to 1, (1-\cos x)/x^2 \to 1/2,$ the second of which follows immediately from the first. $\endgroup$ – zhw. Jul 11 '17 at 6:02
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multiplying numerator and denominator by$$1+\cos(x)$$ we obtain $$\frac{1-\cos(x)^2}{x\sin(x)(1+\cos(x))}=\frac{\sin(x)^2}{x\sin(x)(1+\cos(x))}=\frac{\sin(x)}{x}\cdot \frac{1}{1+\cos(x)}$$

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  • $\begingroup$ First expression should have $1-\cos^2(x)$, right? $\endgroup$ – John Lou Jul 10 '17 at 22:16
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$$\frac{1-\cos x}{x\sin x}=\frac{2\sin^2\frac x2}{2x\sin\frac x2\cos\frac x2}=\frac12\frac{\sin\frac x2}{\frac x2}\frac1{\cos\frac x2}.$$

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hint

Combine the two Well-known limits

$$\lim_{x \rightarrow 0} \frac {\sin (x)}{x}=\lim_{x \rightarrow 0}\frac {x}{\sin (x)}=1$$ and $$\lim_{x \rightarrow 0} \frac {1-\cos (x)}{x^2}=\frac {1}{2} $$

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Use the fact that$$\frac{1-\cos x}{x\sin x}=\frac{(1-\cos x)(1+\cos x)}{x\sin x(1+\cos x)}=\frac{\sin x}{x(1+\cos x)}.$$

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We can also use L'Hospital's Rule

$$\lim_{x\to0}\frac{1-\cos x}{x\sin x}=\lim_{x\to0}\frac{\sin x}{x\cos x+\sin x}=\lim_{x\to0}\frac{\cos x}{-x\sin x+2\cos x}=\frac{1}{2}$$

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Just taking terms of the fraction in question leads to: \begin{align} \frac{1-\cos(x)}{x \, \sin(x)} &= \frac{1 - \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\right)}{x^2 \, \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots \right)} \\ &= \frac{\frac{1}{2!} - \frac{x^2}{4!} + \cdots}{1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots } \\ &= \frac{1}{2!} + \frac{x^2}{4!} + \frac{3 \, x^4}{6!} + \mathcal{O}(x^6) \end{align} which leads to $$\lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} = \frac{1}{2}.$$

Alternatively, L'Hospital's rule applies. \begin{align} \lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} \to \frac{0}{0} \end{align} which leads to \begin{align} \lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{x \, \cos(x) + \sin(x)} \to \frac{0}{0} \end{align} and finally becomes \begin{align} \lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{x \, \cos(x) + \sin(x)} = \lim_{x \to 0} \frac{\cos(x)}{2 \, \cos(x) - x \, \sin(x)} = \frac{\cos(0)}{2 \, \cos(0)} = \frac{1}{2}. \end{align}

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$$\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac{1}{2}\implies 1-\cos(x)\sim\frac{1}{2}x^2\mbox{ for }x\to 0$$

$$\lim_{x\to 0}\frac{\sin(x)}{x}=1\implies \sin(x)\sim x\mbox{ for }x\to 0$$

so $$\lim_{x\to 0}\frac{1-\cos(x)}{x\sin(x)}=\lim_{x\to 0}\frac{\frac{1}{2}x^2}{x\cdot x}=\frac{1}{2}$$

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