Limits of trigonometric functions

Question

I'm stuck on this problem, it seems that I have tried everything in my arsenal but I'm still unable to solve it.. how would you calculate this limit?

I should have added that we are not allowed to use L'Hopital just yet.

Answer 1

You can get tricky with your applications of trig and algebra.

$\frac {\cos 3x - \cos 2x}{x^2}\\ \frac {4\cos^3 x - 3\cos x - 2 \cos^2 x + 1}{x^2}\\ \frac {(\cos x - 1)(4\cos^2 x + 2\cos x - 1)}{x^2}\\ \frac {(\cos x + 1)(\cos x - 1)(4\cos^2 x + 2\cos x - 1)}{(\cos + 1)x^2}\\ \frac {(\cos^2 x - 1)(4\cos^2 x + 2\cos x - 1)}{(\cos + 1)x^2}\\ \frac {(-\sin^2 x) (4\cos^2 x + 2\cos x - 1)}{(\cos + 1)x^2}\\ \lim_\limits{x\to 0} \frac {\sin x}{x} = 1\\ \lim_\limits{x\to 0} \frac {-(4\cos^2 x + 2\cos x - 1)}{(\cos + 1)x^2} = -\frac {5}{2}$

You could apply L'Hopital's rule twice...

$\lim_\limits{x\to 0}\frac {\cos 3x - \cos 2x}{x^2}\\ \lim_\limits{x\to 0}\frac {-3\sin 3x + 2\sin 2x}{2x}\\ \lim_\limits{x\to 0}\frac {-9\cos 3x + 4\cos 2x}{2} = -\frac {5}{2}$

Or, you could use a Taylor series. $\cos x = 1 - \frac {x^2}{2} + o(x^4)$

$\lim_\limits{x\to 0}\frac {\cos 3x - \cos 2x}{x^2}\\ \lim_\limits{x\to 0}\frac {1- \frac {(3x)^2}{2} - 1+ \frac {(2x)^2}{2} + o(x^4)}{x^2}\\ \lim_\limits{x\to 0}-\frac {5}{2} + o(x^2) =-\frac {5}{2}$

Answer 2

Hint :

$$ \dfrac{\cos 3x-\cos 2x}{x^2}=\dfrac{(1-\cos 2x) - (1-\cos3x)}{x^2}$$

Answer 3

Another option uses $\cos(3x)-\cos(2x)=-2\sin(5x/2)\sin(x/2)$, so the limit is $-2(5/2)(1/2)=-5/2$.