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Find the limit where $a$ is a constant $$ \lim_{x \to 0}\frac{\left [ \cos(a+x)-\cos(a-x) \right ]^2}{\tan^2(3x)} $$

I don't know what to do. At first I thought I could replace $a$ with an arbitrary number and then solve the limit but then I got stuck as to how to use the squeeze theorem on this question. Any help?

Michael Hardy
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StrugglingCalcStudent
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4 Answers4

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Hint: $$ \frac{\left[ \cos(a+x)-\cos(a-x) \right]^2}{\tan^2(3x)} = \frac{4\sin^2 a \sin^2 x}{\tan^2(3x)} = \frac{4\sin^2 a }{9}\frac{(3x)^2}{\tan^2(3x)} \frac{\sin^2 x}{x^2}.$$

njguliyev
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  • Is there a trig identity I should be aware of because I can see that you formatted the question so that the limits for two could be found and the limit for 4sin^2(a)/9 depends on a... Is that right? – StrugglingCalcStudent Oct 18 '13 at 21:42
  • Yes, that is right. I used $\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$. – njguliyev Oct 18 '13 at 21:44
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$$ \lim_{x\to0}\frac{\cos(a+x)-\cos a}{x} = \cos'(a) = -\sin a $$ and $$ \lim_{x\to0}\frac{\cos a - \cos(a-x)}{x} = \cos' a = -\sin a, $$ so adding these, we get $$ \lim_{x\to0}\frac{\cos(a+x)-\cos(a-x)}{x} = 2\cos'a = -2\sin a. $$ Hence $$ \lim_{x\to0}\frac{\cos(a+x)-\cos(a-x)}{\tan(3x)} = \lim_{x\to0}\frac{x}{\tan(3x)}\cdot\lim_{x\to0}\frac{\cos(a+x)-\cos(a-x)}{x}. $$ The first limit in the last expression can be found via L'Hopital's rule or by other methods; the second is what we just found.

Finally, square the whole thing.

Michael Hardy
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By the definition of the derivative we have $$\lim_{x\to0}\frac{\cos(a+x)-\cos(a-x)}{x}=(\cos(a+x)-\cos(a-x))'\big|_{x=0}=-2\sin a$$ and

$$\lim_{x\to0}\frac{\tan( 3x)}{x}=(\tan(3 x))'\big|_{x=0}=3$$ hence we can see easily that $$ \lim_{x \to 0}\frac{\left [ \cos(a+x)-\cos(a-x) \right ]^2}{\tan^2(3x)} =\frac{4}{9}\sin^2a$$

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Even if we only know that $S=\lim\limits_{x\to0}\frac{\sin(x)}{x}$ exists, but don't know it's value (other than $S\ne0$), we have $$ \begin{align} \left(\frac{\cos(a+x)-\cos(a-x)}{\tan(3x)}\right)^2 &=\left(\frac{2\sin(a)\sin(x)}{\tan(3x)}\right)^2\\ &=\left(\frac23\sin(a)\cos(3x)\frac{\sin(x)}x\frac{3x}{\sin(3x)}\right)^2\tag{1} \end{align} $$ Taking the limit of $(1)$ as $x\to0$ yields $$ \begin{align} &\lim_{x\to0}\left(\frac{\cos(a+x)-\cos(a-x)}{\tan(3x)}\right)^2\\ &=\frac49\sin^2(a)\lim_{x\to0}\cos^2(3x)\left(\lim_{x\to0}\frac{\sin(x)}x\right)^2\left(\lim_{x\to0}\frac{\sin(3x)}{3x}\right)^{-2}\\ &=\frac49\sin^2(a)\cdot1\cdot S^2\cdot\frac1{S^2}\\ &=\frac49\sin^2(a)\tag{2} \end{align} $$

robjohn
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