Find the limit of a function as $ x$ approaches $0$

Question

How can I find the limit of $\dfrac{\cos(3x) - 1 }{x^2}$ as $x\to 0$?

If someone could please break down the steps, for clear understanding. I'm studying for the GRE. Thanks in advance !!

Answer 1

Since you are studying for the GRE, which if I recall correctly is a multiple choice exam, knowing L'Hospital's rule is a good thing. Here's the simple version:

When you plug in your value that $x$ is approaching, if you end up with $0/0$ or $\infty/\infty$, then you can take the derivatives of the top and bottom, and then take the limit again. Using this rule (which we have to apply twice here),

$$\lim\limits_{x\to 0}\dfrac{\cos(3x)-1}{x^2}=\lim\limits_{x\to0}\dfrac{-3\sin(3x)}{2x}=\lim\limits_{x\to0}\dfrac{-9\cos(3x)}{2},$$

and then plugging in $0$ gives our answer, which is $-9/2$. Does this all make decent sense?

Answer 2

It can be done without L'hospitale formula as follows:

a) Use the formula $1-\cos(3x) = 2\sin^2(\frac{3x}{2})$

b) Use the formula $\dfrac{\sin x}{x} \to 1$ when $x \to 0$

Answer 3

By Taylor Expansion:

$$\lim_{x\to 0}\frac{\cos(3x)-1}{x^2}=\lim_{x\to 0}\frac{(1-9x^2/2+\cdots) - 1}{x^2}\stackrel{x\to 0} = -\frac{9}{2}$$