MHB PBM.1 Limit to Zero: $$\lim_{x\to 0} \frac{\cos 3x-1}{x^2}$$

[karush]

$$\lim_{{x}\to{0}}\frac{\cos\left({3x}\right)-1}{{x}^{2}}$$
$$\frac{f'}{g'}
=-\frac{3\sin\left({3x}\right)}{2x}
=-\frac{9}{2}\cdot\frac{\sin\left({3x}\right)}{3x}$$
$x\to 0$ is $-\frac{9}{2}$

Just seeing if this is correct or better way to do it

 

[MarkFL]

That's correct, but if you are allowing yourself L'Hôpital, then I would just write:

$$L=\lim_{x\to0}\frac{\cos(3x)-1}{x^2}=-\frac{3}{2}\lim_{x\to0}\frac{\sin(3x)}{x}=-\frac{9}{2}\lim_{x\to0}\cos(3x)=-\frac{9}{2}$$

 

[Greg]

Correct! Well done!

 

[Greg]

$$\lim_{x\to0}\dfrac{\cos3x-1}{x^2}$$

$$x=\dfrac13y$$

$$\lim_{y\to0}\dfrac{9(\cos y-1)}{y^2}$$

$$\lim_{y\to0}\dfrac{-9\sin^2y}{y^2(\cos y+1)}=-9\cdot1\cdot\dfrac12=-\dfrac92$$

 

[karush]

Sorry but I don't see the purpose of sticking $y$ in this thing

 

[MarkFL]

Sorry but I don't see the purpose of sticking $y$ in this thing

Yes, you could do it without making a substitution, but it just looks cleaner if you do. However, the main takeaway from Greg's post is how to take the limit without using L'Hôpital's Rule at all. :)