# Homework Help: Finding lim of following question

1. Jan 4, 2005

### KataKoniK

I am having trouble finding the lim of the following question. Any help?

lim
theta -> 0+

(theta - sin theta) / 2 ( tan (1/2)theta - 1/2 theta)

The answer is suppose to be 2, but I keep on getting 1.

2. Jan 4, 2005

### Hurkyl

Staff Emeritus
What have you done?

3. Jan 4, 2005

### KataKoniK

I have done the following, but don't know where to go afterwards.

1 - cos theta / sec^2 (1/2) theta
= 1 - cos theta / 1 + tan^2 (1/2) theta

However, it's not in the form 0/0, so i can't apply L'Hopital anymore. Any idea?

4. Jan 4, 2005

### marlon

you mean $$\frac{\theta - sin(\theta)}{2(tan(\frac{\theta}{2}) - \frac{\theta}{2})}$$

If so, just apply de l'Hôpital two times in a row and you case is done...

regards
marlon

5. Jan 4, 2005

### Zurtex

It may not be in the form 0/0 but can't you let theta=0 anyway? Also in the line above couldn't you apply the fact that sec theta = 1 / cos theta?

I don't know I've never really had to take the limits of anything.

6. Jan 4, 2005

### dextercioby

Applying 2 times the rule of Antoine L'Ho^spital very,very carefully,u get '2'.Trust me.

An elegant alternative proof (i like it more than L'Ho^spital):
$$\lim_{\theta\searrow 0} \frac{\theta-\sin\theta}{2(\tan\frac{\theta}{2}-\frac{\theta}{2})}$$(1)

Since the limits is to 0,we know for about 300 years that,close to zero
$$\sin\theta\sim \theta+\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!}$$(2)
$$\cos\theta\sim 1+\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}$$(3)

From (2) and (3) i use only the first 2 terms.I plug these expressions into (1),after expressing '\tan' as a ratio between 'sin' and 'cos'.
$$\lim_{\theta\searrow 0} \frac{\theta-\sin\theta}{2(\tan\frac{\theta}{2}-\frac{\theta}{2})}=\lim_{\theta\searrow 0}\frac{\theta-\theta-\frac{\theta^{3}}{3!}}{2[\frac{\theta^{3}}{8\cdot 3!}-\frac{\theta}{2}(1+\frac{\theta^{2}}{4\cdot 2!})]}\cos\frac{\theta}{2}$$
$$=\lim_{\theta\searrow 0}\frac{-\frac{\theta^{3}}{3!}}{2(\frac{\theta^{3}}{8\cdot 3!}-\frac{\theta^{3}}{8\cdot 2!})}\cos\frac{\theta}{2}$$(4)

Simplify through the function at the numerator and find
$$\lim_{\theta\searrow 0}\frac{\theta-\sin\theta}{2(\tan\frac{\theta}{2}-\frac{\theta}{2})}=\lim_{\theta\searrow 0}\frac{1}{2(-\frac{1}{8}+\frac{3}{8})}\cos\frac{\theta}{2}=\frac{8}{4}=2$$(5)

Daniel.

PS.Math is beauty...

7. Jan 4, 2005

### Hurkyl

Staff Emeritus
I assume you mean that you applied L'Hôpital's rule to:

$$\frac{\theta - \sin \theta}{2(\tan (1/2)\theta - (1/2)\theta)}$$

and got

$$\frac{1 - \cos \theta}{\sec^2 (1/2)\theta}$$

First comment: How did you get 1? This limit goes to 0 as &theta; approaches 0. (numerator goes to 0, denominator goes to 1)

Second comment: For some reason, you know your answer is wrong. Well, you've only done three things:

(1) Decided to apply L'Hôpital's rule
(2) Differentiated the numerator
(3) Differentiated the denominator

Logic dictates that you made a mistake in one of these...

P.S. please use parentheses around your numerators and denominators. In proper usage, 1 + 2 / 3 + 4 means $1 + \frac{2}{3} + 4$, not $\frac{1 + 2}{3 + 4}$.

Last edited: Jan 4, 2005
8. Jan 5, 2005

### KataKoniK

Got it. Thanks for your help.