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

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?

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

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

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

Hint:

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

• Digestible hint @Mehenni +=) Commented Sep 20, 2013 at 8:59
• @BabakS.: Thanks for the comment. How are you doing by the way? Commented Sep 20, 2013 at 9:00
• 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! Commented Sep 20, 2013 at 9:21
• @hejseb: You are welcome. By the way, there is a nice geometric proof for $\lim_{x\to 0}\frac{\sin(x)}{x}=1$. Commented Sep 20, 2013 at 9:45
• @MhenniBenghorbal: I am just ready to start a new academic year. A new teaching period. :-) Commented Sep 20, 2013 at 9:53

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