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I can't figure out how to get the limit in this problem. I know that ${1-\cos x \over x}=0$ but I'm not allowed to use L'Hopital's Rule. I also already know that the answer is $-{25 \over 36}$ but I don't know the steps in between. I've already tried multiplying by the conjugates of both the numerator and the denominator but neither are getting me anywhere close. Here is the question:

$$\lim_{x\to 0}{1-\cos 5x \over \cos 6x-1}$$

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  • $\begingroup$ Are you allowed to use $\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$? The exercise becomes very easy using it. $\endgroup$ – Federico Poloni Sep 30 '14 at 6:40
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Outline: Our expression is equal to $$-\frac{1+\cos 6x}{1+\cos 5x}\cdot\frac{1-\cos^25x}{1-\cos^2 6x},\tag{1}$$ which is $$-\frac{1+\cos 6x}{1+\cos 5x}\cdot\frac{\sin^2 5x}{\sin^2 6x}.\tag{2}$$ To find $$\lim_{x\to 0} \frac{\sin 5x}{\sin 6x},$$ rewrite as $$\frac{5}{6}\lim_{x\to 0} \frac{\frac{\sin 5x}{5x}}{\frac{\sin 6x}{6x}}.$$

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  • $\begingroup$ thanks for your answer! could you explain the first two steps a little more? and where is the negative sign that should be in the answer? $\endgroup$ – Emeraldis Sep 30 '14 at 3:09
  • $\begingroup$ I wanted the top to be $1-\cos^2 5x$. So I multiplied top and bottom by $1+\cos 5x$. I wanted the bottom to be $1-\cos^2 6x$. So I pulled a $-1$ to the front, and multiplied top and bottom by $1+\cos 6x$. That gets us to (1). To get to (2), use the fact that $1-\cos^2 t=\sin^2 t$. The $\frac{1+\cos 6x}{1+\cos 5x}$ in front of (2) approaches $1$ as $x\to 0$. So we are only worried about the $\frac{\sin^2 5x}{\sin^2 6x}$ part. I dealt with $\frac{\sin 5x}{\sin 6x}$ in the second part of the post, leaving it to you to put the pieces together. If there are more questions, please leave a message. $\endgroup$ – André Nicolas Sep 30 '14 at 3:26
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Hint. First, note that $$\lim_{x\to0}\frac{\sin ax}{x}=a\ ,$$ which can be proved by geometric arguments without using L'Hopital's rule.

We have $$\eqalign{\frac{1-\cos ax}{x^2} &=\frac{1-\cos ax}{x^2}\frac{1+\cos ax}{1+\cos ax}\cr &=\Bigl(\frac{\sin ax}{x}\Bigr)^2\frac{1}{1+\cos ax}\ .\cr}$$ You should now be able to find the limit of this expression as $x\to0$, and hence solve your problem.

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You could also use Taylor series since $$\cos(y)=1-\frac{y^2}{2}+O\left(y^3\right)$$ Now replace successively $y$ by $5x$ and get the numerator $$1-\cos(5x)=\frac{25 x^2}{2}+O\left(x^3\right)$$ and get the denominator replacing $y$ by $6x$, so $$\cos(6x)-1=-18 x^2+O\left(x^3\right)$$ So the ratio is what you said.

If you use more terms for each development and perform long division, you should end with $${1-\cos (5x) \over \cos (6x)-1}=-\frac{25}{36}-\frac{275 x^2}{432}-\frac{1595 x^4}{2592}+O\left(x^5\right)$$ which shows how the limit is approached when $x$ goes to $0$.

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