# Trig limit without using l'hopital's rule [closed]

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

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

## closed as off-topic by user370967, B. Mehta, Dave, samjoe, Xander HendersonOct 23 '17 at 19:17

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• I don't get this. This question clearly shows no effort, and other questions get closed for this reason, but this one gets upvotes and answers? – user370967 Oct 23 '17 at 14:18
• There should be a new flag for zero effort shown by OP in a text book type problem. – Narasimham Oct 23 '17 at 18:31

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.

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

• How did we get to $3x$ in $\frac{sin3x}{3x}$? – gbox Oct 23 '17 at 7:07
• simple: write $x = 3\left(\dfrac{x}{3}\right)$ – DeepSea Oct 23 '17 at 7:08

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.

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