Apply L'Hopital's rule (LHR) two times:
[tex]\displaystyle \lim_{x\to0} \frac{\cos^2(x) - 1}{x^2} \stackrel{LHR}= \lim_{x\to0} \frac{-2\cos(x) \sin(x)}{2x} \stackrel{LHR}= \lim_{x\to0}\frac{2\sin^2(x) - 2\cos^2(x)}{2}[/tex]
Recall that
cos(2x) = cos²(x) - sin²(x)
So our limit is equivalent to
[tex]\displaystyle \lim_{x\to0}\frac{2\sin^2(x)-2\cos^2(x)}2 = \lim_{x\to0} -\cos(2x) = -\cos(0) = \boxed{-1}[/tex]
You can also do this easily without LHR, provided that you know
[tex]\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1[/tex]
We again have
[tex]\displaystyle \lim_{x\to0}\frac{\cos^2(x)-1}{x^2} = \lim_{x\to0}\frac{-\sin^2(x)}{x^2} = -\lim_{x\to0}\left(\frac{\sin(x)}x\right)^2 \\\\ = -\left(\lim_{x\to0}\frac{\sin(x)}x\right)^2 = -1^2 = -1[/tex]
