Finding $\lim_{x\to 0}\dfrac{\cos x-1}{x^3}$ I understand that in order for the limit to exist, the limit as $x$ approaches from the left of 0 has to be equal to the limit as $x$ approaches from the right of 0. In this case, I found that one leads to infinity and the other leads to negative infinity. However, the answer from the book claims that the limit is negative infinity.
 A: Hint
$$\lim_{x\to 0}\frac{\cos x -1}{x^3}=\lim_{x\to 0}\left(-\frac{1}{x}\right)\left(\frac{1}{\cos x+1}\right)\left(\frac{\sin x}{x}\right)^2.$$
I think you are right.
A: Yeah so the book is wrong and you are right. Indeed first notice that
$$\left\lvert\frac{\cos x-1}{x^3}\right\rvert\geq\left\lvert\frac{x^2}{2x^3}\right\rvert=\frac{1}{2\lvert x\rvert}\to\infty.$$
Furthermore, notice that
$$\cos x-1\leq 0,$$
so the sign of $\frac{\cos x-1}{x^3}$ is the opposite of the sign of $x^3$, and so in particular,
$$\frac{\cos x-1}{x^3}\geq 0,\quad x\in(-\infty,0),$$
$$\frac{\cos x-1}{x^3}\leq 0,\quad x\in(0,\infty).$$
Combining these facts we get that
$$\lim_{x\to0^+}\frac{\cos x-1}{x^3}=-\infty$$
while
$$\lim_{x\to0^-}\frac{\cos x-1}{x^3}=\infty,$$
and so as the one-sided limits do not coincidence, the limit is not equal to $-\infty$.
A: You can apply L'Hopital's rule twice to settle the question.
$$\lim_{x\to 0} \frac{\cos(x) - 1}{x^3} = \lim_{x\to 0} \frac{-\sin(x)}{3x^2} = \lim_{x\to 0} \frac{-\cos(x)}{6x}. \tag1 $$
As $x$ goes to $(0^-)$ (i.e. approaches $(0)$ from below), the far right fraction in (1) above indicates that the (left side) limit is $$\frac{-1}{- ~(0)} = +\infty.$$
As $x$ goes to $(0^+)$ (i.e. approaches $(0)$ from above), the far right fraction in (1) above indicates that the (right side) limit is $$\frac{-1}{+ ~(0)} = -\infty.$$
