Question

Resuelve con la regla de L’Hopital

Answer 1

Apply L'Hopital's rule (LHR) two times:

`\displaystyle \lim_(x\to0) (\cos^2(x) - 1)/(x^2) \stackrel{LHR}= \lim_(x\to0) (-2\cos(x) \sin(x))/(2x) \stackrel{LHR}= \lim_(x\to0)(2\sin^2(x) - 2\cos^2(x))/(2)`

Recall that

cos(2x) = cos²(x) - sin²(x)

So our limit is equivalent to

`\displaystyle \lim_(x\to0)\frac{2\sin^2(x)-2\cos^2(x)}2 = \lim_(x\to0) -\cos(2x) = -\cos(0) = \boxed{-1}`

You can also do this easily without LHR, provided that you know

`\displaystyle \lim_(x\to0)\frac{\sin(x)}x = 1`

We again have

`\displaystyle \lim_(x\to0)(\cos^2(x)-1)/(x^2) = \lim_(x\to0)(-\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`