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How do I find the following limit:

$$ \lim_{x\to 0}\frac{\tan x(1- \cos x)}{3x^2} $$

without using L'Hopital's rule? The reason I'm making a point of not using L'Hopital is that if I run the limit through Wolfram Alpha that's the method it uses, but we haven't gone through that yet so I'm guessing I should use something else.

I don't really know what to do here. I've done quite a number of exercises on limits by now and I almost always get it right and know immediately what to do, but not this time. The only thing I can think of is to use $\tan x=\frac{\sin x}{\cos x}$. But I can't say that helps me much... That only gives me $$ \lim_{x\to 0}\frac{\tan x(1- \cos x)}{3x^2}=\lim_{x\to 0}\frac{\sin x(\cos^{-1}x- 1)}{3x^2} $$

Does that help me? What should I do next? Or should I start with something different?

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  • $\begingroup$ Have you tried Taylor expansion of the numerator? $\endgroup$
    – Daniel R
    Commented Sep 20, 2013 at 8:52
  • $\begingroup$ @DanielR No, because I don't know that unfortunately. $\endgroup$
    – hejseb
    Commented Sep 20, 2013 at 8:53

3 Answers 3

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Rewriting the limit in a manner similar to what you were doing, we find that

$$\frac{\tan{x} (1 - \cos{x})}{3x^2} = \frac{\sin{x}}{x} \frac{1 - \cos{x}}{x} \frac{1}{3\cos{x}}$$

The first and third terms are bounded, while the second term tends to zero. One way to see this is to multiply it top and bottom by $1 + \cos{x}$; this gives

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

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Hint:

$$1-\cos(x) = 2\sin(x/2)^2. $$

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  • $\begingroup$ Digestible hint @Mehenni +=) $\endgroup$
    – Mikasa
    Commented Sep 20, 2013 at 8:59
  • $\begingroup$ @BabakS.: Thanks for the comment. How are you doing by the way? $\endgroup$
    – Mhenni Benghorbal
    Commented Sep 20, 2013 at 9:00
  • $\begingroup$ It took me a while to get this one! But using this form I get $\lim_{x\to 0}\frac{2\tan x}{3}\left(\frac{\sin\frac{x}{2}}{x}\right)^2=\frac{1}{4}\lim_{x\to 0}\frac{2\tan x}{3}=\frac{2\cdot 0}{12}=0$. Now I only need to show why the limit of $\frac{\sin\frac{x}{2}}{x}$ is 1/2, but that's a bit easier! Thanks! $\endgroup$
    – hejseb
    Commented Sep 20, 2013 at 9:21
  • $\begingroup$ @hejseb: You are welcome. By the way, there is a nice geometric proof for $\lim_{x\to 0}\frac{\sin(x)}{x}=1$. $\endgroup$
    – Mhenni Benghorbal
    Commented Sep 20, 2013 at 9:45
  • $\begingroup$ @MhenniBenghorbal: I am just ready to start a new academic year. A new teaching period. :-) $\endgroup$
    – Mikasa
    Commented Sep 20, 2013 at 9:53
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Series expansion. $\tan(x)$ is starting with first power, $1-\cos(x)$ is second power, so the expression is of form $a x + b x^3 +...$ where $a$ and $b$ are constants. Goes to zero in limit.hth

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