# Evaluating $\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}$

The problem is, as stated:

$$\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}$$ Of course, a and b are real numbers.

I tried implementing the trig identity:

$$\cos(ax)-\cos(bx) = -2\sin\frac{(ax-bx)}{2}\sin\frac{(ax+bx)}{2}$$

But that didn't really take me anywhere. In my book, similar problems were often solved this way but here that doesn't seem to work.

One obviously has to try and use the known limit to reduce this to a more easy limit:

$$\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$$

Any help would be much appreciated.

• Have you learned L'Hopitals Rule? Mar 22, 2018 at 15:01
• Yes, but this particular problem was part of the set that was supposed to be solved without using it. Mar 22, 2018 at 15:03

The identity given in your question is useful for evaluating the limit: as $x\to 0$ we have $$\frac{\cos(ax) - \cos(bx)}{\sin^2(x)} = -2\frac{\frac{\sin\frac{(ax-bx)}{2}}{x}\cdot \frac{\sin\frac{(ax+bx)}{2}}{x}}{\frac{\sin x}{x}\cdot\frac{\sin x}{x}}\to -2\frac{\frac{a-b}{2}\cdot \frac{a+b}{2}}{1\cdot 1}=\frac{b^2-a^2}{2}$$ where we used the fact that for any real $\alpha$, $$\lim_{x\to 0}\frac{\sin(\alpha x)}{x}=\alpha\lim_{x\to 0}\frac{\sin(\alpha x)}{(\alpha x)}=\alpha\cdot 1=\alpha.$$

$$\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}=\lim_{x\rightarrow 0}\frac{x^2}{\sin^2(x)}\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{x^2}\\=-\lim_{x\rightarrow 0}\frac{1- \cos(ax)}{x^2}+\lim_{x\rightarrow 0}\frac{1- \cos(bx)}{x^2}=\frac{ b^2-a^2}{2}$$

Given that $$\lim_{x\rightarrow 0}\frac{1- \cos(x)}{x^2}=1/2$$ $$\lim_{x\rightarrow 0}\frac{ \sin(x)}{x}=1.$$

Hint For Finishing The Approach In The Question $$\frac{\cos(ax)-\cos(bx)}{\sin^2(x)}=2\frac{\sin\left(\frac{b-a}2\,x\right)}{\sin(x)}\frac{\sin\left(\frac{b+a}2\,x\right)}{\sin(x)}$$

Hint For Another Approach \begin{align} \frac{\cos(ax)-\cos(bx)}{\sin^2(x)} &=\frac{1-\cos(bx)}{\sin^2(x)}-\frac{1-\cos(ax)}{\sin^2(x)}\\ &=\frac1{1+\cos(bx)}\frac{\sin^2(bx)}{\sin^2(x)}-\frac1{1+\cos(ax)}\frac{\sin^2(ax)}{\sin^2(x)} \end{align}

\begin{align}\lim_{x\to0}\frac{\cos(ax)-\cos(bx)}{\sin^2x}&=\lim_{x\to0}\frac{x^2}{\sin^2x}\cdot\frac{\cos(ax)-\cos(bx)}{x^2}\\&=\lim_{x\to0}\frac{\cos(ax)-\cos(bx)}{x^2}\\&=\frac{b^2-a^2}2\end{align}since$$\cos(ax)-\cos(bx)=-\frac12(a^2-b^2)x^2+o(x^2).$$

• I think your answer is a little wrong. Can you please check again? Mar 22, 2018 at 15:05
• How did you get that last identity? Mar 22, 2018 at 15:05
• @JaideepKhare There was indeed an error, which I have corrected. Mar 22, 2018 at 15:07
• Shouldn't it be $b^2-a^2 \over 2$? Mar 22, 2018 at 15:11
• @Lorenzo I've edited my answer. Thank you. Mar 22, 2018 at 15:32

You are on the right track,

$$\lim_{x\rightarrow 0} \frac{\cos (ax)-\cos (bx)}{\sin^2 x}=\lim_{x\rightarrow0}\frac{-2 \sin \left(\frac{a+b}{2}x \right)\sin \left(\frac{a-b}{2}x \right)}{\sin^2 x}=\lim_{x\rightarrow0}-2\frac{\frac{\sin \left(\frac{a+b}{2}x \right)}{x}}{\frac{\sin x}{x}}\frac{\frac{\sin \left(\frac{a-b}{2}x \right)}{x}}{\frac{\sin x}{x}}.$$

It should be obvious now.

You can use L' Hospital rule $$\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}$$ $$=\lim_{x\rightarrow 0}\frac{-a\sin(ax) +b \sin(bx)}{2\sin(x)\cos(x)}$$ $$=\lim_{x\rightarrow 0}\frac{-a^2\cos(ax) +b^2 \cos(bx)}{2\cos^2(x)-2\sin^2(x)}=\frac{b^2-a^2}{2}$$ Where we applied L'Hospital's rule twice

• @farruhota ah yes, I didn’t notice that mistake Mar 22, 2018 at 16:17

i would write $$\frac{-2\left(\sin x\left(\frac{a-b}{2}\right)\right)\sin\left(x\left(\frac{a+b}{2}\right)\right)x^2\left(\frac{a^2-b^2}{4}\right)}{x\left(\frac{a-b}{2}\right) x\left(\frac{a+b}{2}\right)\sin^2(x)}$$

I believe this approach is simpler: since $1-\cos(cx)=2\sin^2\frac{cx}{2}$ and $\lim_{x\to 0}\frac{\sin(dx)}{\sin(x)}=d$,

$$\lim_{x\to 0}\frac{\cos(ax)-\cos(bx)}{\sin^2(x)}=\lim_{x\to 0}\frac{2\sin^2\frac{bx}{2}-2\sin^2\frac{ax}{2}}{\sin^2(x)}=2\left[\left(\frac{b}{2}\right)^2-\left(\frac{a}{2}\right)^2\right]=\color{blue}{\frac{b^2-a^2}{2}}.$$

From the Taylor series of $\sin(x),\cos(x)$ and using $O$-notation, we know that, for $x\to 0$, \begin{eqnarray*} \sin x &=&x+O\left( x^{3}\right) \\ \cos x &=&1-\frac{1}{2}x^{2}+O\left( x^{3}\right). \end{eqnarray*}

Then $$\sin ^{2}x=x^{2}+O\left( x^{4}\right)$$ and $$\frac{\cos ax-\cos bx}{\sin ^{2}x}=\frac{\left( -\frac{1}{2}a^{2}+\frac{1}{2} b^{2}\right) x^{2}+O\left( x^{3}\right) }{x^{2}+O\left( x^{4}\right) }.$$ Hence $$\lim_{x\rightarrow 0}\frac{\cos ax-\cos bx}{\sin ^{2}x}=\lim_{x\rightarrow 0} \frac{x^{2}\left( -\frac{1}{2}a^{2}+\frac{1}{2}b^{2}\right) }{x^{2}}=\frac{1 }{2}b^{2}-\frac{1}{2}a^{2}.$$