Trigonometric limit $\lim_{x\to 0} \frac{\cos(a+2x) - 2\cos(a+x) + \cos(a)}{x^2}$ without l'Hospital

Question

Compute the following limit:

$$\lim_{x\to 0} \frac{\cos(a+2x) - 2\cos(a+x) + \cos(a)}{x^2}\ .$$

How would I go about solving this problem? I can't use l'Hospital.

Answer 1

Simply expand: $$ \cos(a+2x) = \cos a \cos(2x) - \sin a\sin(2x) = 2\cos a \cos^2 x - \cos a - 2\sin a\sin x \cos x \\ \cos(a+x) = \cos a \cos x - \sin a\sin x $$ Putting things together, you find $$ \frac{ 2\cos a \cos^2 x - 2\sin a\sin x \cos x - 2 \cos a \cos x + 2 \sin a\sin x }{x^2} $$ Now use the fact that $$ \frac{1-\cos u}{u^2}\to \frac 12 $$to conclude.

Answer 2

$$\cos (a+2x) -2\cos (a+x) +\cos a = \cos (a+2x) -\cos (a+x) -(\cos (a+x) -\cos a) =-2\sin \left(a+\frac{3x}{2}\right)\sin\frac{x}{2} +2\sin \left(a+\frac{x}{2} \right)\sin\frac{x}{2} =-4\sin\frac{x}{2}\sin\frac{x}{2}\cos (a+x) $$ hence $$\lim_{x\to 0} \frac{\cos (a+2x) -2\cos (a+x) +\cos a}{x^2} =\lim_{x\to 0}\frac{-4\sin^2\frac{x}{2} \cos(x+a)}{x^2} =-\cos a$$

Answer 3

Let $m = a+x.$ Then $$\begin{eqnarray} \cos(a+2x) - 2\cos(a+x) + \cos(a) &=& \cos(m+x) - 2\cos m + \cos(m-x) \\ &=& \cos m \cos x - \sin m \sin x - 2\cos m + \cos m \cos x + \sin m \sin x \\ &=& 2 (\cos m)(\cos x - 1) \\ &=& 2 \cos(a+x)(\cos x - 1). \end{eqnarray}$$ You are therefore trying to find $$\lim_{x\to 0} \left(2 \cos(a+x) \frac{\cos x - 1}{x^2}\right) = 2 \left(\lim_{x\to 0} \cos(a+x)\right)\left(\lim_{x\to 0} \frac{\cos x - 1}{x^2}\right).$$