What is the limit without using l'Hopital's rule.

Question

$$ \lim_{x \to 0} \frac{1-\cos x}{x\sin 2x} $$

So I was thinking of separating it into two parts, one is $(1-\cos(x))/x$ and the other one is $1/\sin(2x)$. The limit for the first part is 0, but I don't know what to do for the second part. Am I approaching it correctly in the first place? Any hint will be great!

Answer 1

If you decompose into Maclaurin series, you get $\cos x \approx 1 - x^2/2, \sin (2x) \approx 2x$. Can you use this to approximate the limit?

Answer 2

Hint:

$\dfrac{1-\cos(x)}{x\sin(2x)}=$

$\dfrac{1-\cos^2(x)}{2x\sin(x)\cos(x)(1+\cos(x))}=$

$\dfrac{\sin^2(x)}{2x\sin(x)\cos(x)(1+\cos(x))}=$

$\dfrac{\sin(x)}{x}\dfrac{1}{2\cos(x)(1+\cos(x))}.$

Can you take it from here?

Answer 3

If you're allowed to use the fact that $\frac{\sin x}{x} \rightarrow 1$ as $x \rightarrow 0,$ then you can get the limit by multiplying both the numerator and the denominator by $1 + \cos x$ and using a couple of basic trig. identities:

$$ \frac{1 - \cos x}{x \sin {2x}} \cdot \frac{1 + \cos x}{1 + \cos x} \;\; = \;\; \frac{1 - \cos^2 x}{x \cdot \sin {2x} \cdot (1 + \cos x)} \;\; = \;\; \frac{\sin^2 x}{x \cdot 2 \sin x \cos x \cdot (1 + \cos x)}$$

$$ = \;\; \frac{\sin x}{2x \cos x (1 + \cos x)} \;\; = \;\; \frac{1}{2} \cdot \frac{\sin x}{x} \cdot \frac{1}{\cos x (1 + \cos x)} \;\; \longrightarrow \;\; \frac{1}{2} \cdot 1 \cdot \frac{1}{1\cdot 2} \;\; = \;\; \frac{1}{4} $$

Regarding your later-added comment, pretty much every trig limit you'll encounter that you're supposed to evaluate without using L'Hopital's Rule will make use of $\lim \limits_{x\rightarrow 0}\frac{\sin x}{x} = 1$ after some algebraic manipulation (which may involve using trig conjugates, which are analogous to the conjugates you sometimes use when square roots are involved) and the use of trig identities. Although many of the manipulations involved may seem a bit tricky, after a bit of practice you'll get to the point where for most problems of this type there will only be a few obvious things to try, and for textbook and class test problems, one of those obvious things is going to work.