Trig limit without using l'hopital's rule

Question

I need help calculating this limit without L'Hopital's rule.

$$\lim_{x \to 0} \frac{\cos(6x)-1}{x\sin(6x)}$$

Answer 1

hint: Use the followings facts:

$1)$ $\cos(6x) - 1 = -2\sin^2(3x)$,

$2)$ $\sin (6x) = 2\sin(3x)\cos(3x)$,

$3)$ $\sin(3x)/3x \to 1$ for $x \to 0$ of course.

$4)$ $\cos(3x) \to 1$.

Answer 2

The fast way:

$$\cos(6x)-1=-2\sin^2(3x)\sim -18x^2,\\x\sin(6x)\sim6x^2$$

hence $-3$.

---

This is easily made rigourous by using a few $\dfrac{\sin t}t$ ratios.

Answer 3

Multiply numerator and denominator by $6(\cos(6x)+1)$, to get $$ \frac{\cos^2(6x)-1}{6x\sin(6x)}\frac{6}{\cos(6x)+1}= \frac{-\sin^2(6x)}{6x\sin(6x)}\frac{6}{\cos(6x)+1}= -\frac{\sin(6x)}{6x}\frac{6}{\cos(6x)+1} $$