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

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!

gt6989b
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Hatem Chalak
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  • The limit of the second as $x \to 0^+$ is $\infty$, so you get an indeterminate form $0 \cdot \infty$ which doesn't help. – Daniel Schepler Oct 09 '17 at 20:55
  • Even though you're not supposed to use l'Hopital in the solution, you can use it to "cheat" and see what the limit ought to be with l'Hopital, then from there prove that that's the limit without l'Hopital. – Arthur Oct 09 '17 at 20:56
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    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 the numerator and denominator by $1 + \cos x$ and using a couple of basic trig. identities. – Dave L. Renfro Oct 09 '17 at 20:58

3 Answers3

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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.

Dave L. Renfro
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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?

Peter Szilas
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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?

gt6989b
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  • That's no different from l'Hopital, though. – Arthur Oct 09 '17 at 21:07
  • @Arthur (1) OP has not indicated not to use L'Hospital. (2) it is in similar spirit but is somewhat of a different mechanism. – gt6989b Oct 09 '17 at 21:10
  • (1) Did you read the title? Or the tags? (2) the mechanism is identical if you get the series expansion by differentiating and evaluating at $x=0$. The theory behind may differ slightly, but the calculations are the same. – Arthur Oct 09 '17 at 21:19
  • They do not let us use techniques we did not learn yet (even if we know them), so I cannot use L'Hopital's rule or series to find limits. I am interested in learning about the Maclaurin series now though – Hatem Chalak Oct 09 '17 at 21:22
  • Do you know equivalence of functions? – Bernard Oct 09 '17 at 21:26
  • @HatemChalak Maclaurin series are basically the same as Taylor series, and I think it's easier to find material if you search for that. The idea is that nice enough functions (including most of the ones you know the name of) are uniquely determined by all their higher order derivatives at a single point (they are called analytic). For instance, among these analytic functions, $\sin x$ is the only one that at $x=0$ has value $0$, first derivative $1$, second derivative $0$, third derivative $-1$, and so on. – Arthur Oct 09 '17 at 21:37
  • @Arthur the tag was not there when I originally read the question. Same calculations does not mean same technique... – gt6989b Oct 09 '17 at 22:41