# : Limits

1. Mar 28, 2007

URGENT: Limits

1. The problem statement, all variables and given/known data
PART A
Let f(x) be a function defined on an open interval containing c, and let L be a real number. Given:

$$\mathop {\lim }\limits_{x \to a} f\left( x \right) = L$$

For each real $$\varepsilon > 0$$ , there exists a real $$\delta > 0$$ such that for all x:
$$0 < \left| {x - c} \right| < \delta 0 < \left| {f\left( x \right) - L} \right| < \varepsilon$$

PART B
Hence, evaluate exactly $$\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta }$$

2. The attempt at a solution
I do not have the slightest idea on how to solve part A, however for part B my working is as follows (I am using L'hopital's Rule):
$$\begin{array}{l} \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta } = \frac{{\sin ^2 4\left( 0 \right)}}{0} = \frac{0}{0} \\ \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta } = \mathop {\lim }\limits_{\theta \to 0} \frac{{f\left( \theta \right)}}{{g\left( \theta \right)}}\quad \Rightarrow \quad f\left( \theta \right) = \sin ^2 4\theta \quad g'\left( \theta \right) = 1 \\ f\left( \theta \right) = \sin ^2 4\theta \\ \sin ^2 \theta = \frac{{1 - \cos 2\theta }}{2} \\ f\left( \theta \right)\sin ^2 4\theta = \frac{{1 - \cos 8\theta }}{2} \\ \end{array}$$

Am i doing this part correctly? Is there an easier way? Many thanks to all help and suggestions offered.

Last edited: Mar 28, 2007
2. Mar 28, 2007

### Gib Z

Yes you are doing it correctly, but instead of converted the sin squared using double angle formulae, just differentiate as it is. I think that should be easier.

3. Mar 28, 2007

### FunkyDwarf

As far as i can see part A has no question to solve, it simply states the definition of the limit

4. Mar 28, 2007

### Gib Z

I think it is asking to prove that statement it provided or something

5. Mar 28, 2007

### jing

Is it not possible to use small angles $${\sin \theta }\approx\theta$$

so $$\frac{{\sin ^2 4\theta }}{\theta }\approx16\theta$$ as $$\theta \rightarrow 0$$

6. Mar 28, 2007

### Gib Z

Well my graphing program says the limit is zero..

7. Mar 28, 2007

### chickens

sin^2 4@ == (sin 4@)^2

then if u do like normal differentiation, u get 8sin4@cos4@
and with the trigo identity, u will get 4 sin8@
thus, that result would approaches 0 for @ --> 0

ps: sorry, dont know how to use the mathematical notation thingy

Last edited: Mar 28, 2007
8. Mar 28, 2007

### sutupidmath

yeah this limit is zero. there is another way, shorter one and simpler one i guess than chickens suggested, although you get the same result. try to multiply both the numerator and denominator by (16@)/(16@), what do you get? then write the numerator as (4*@)^2, so you will get (sin(4@)/4@)^2, and the limit of this is 1, but we have still the limit of 16@ as @->0 so the result is 0.
sorry for my symbols, i hope u understand @ stands for theta, the angle at sin

9. Mar 28, 2007

### cristo

Staff Emeritus
Can you explain this further, as I don't really understand what you mean! Are you doing this? $$\lim_{x\rightarrow 0} \frac{\sin^24\theta}{\theta}=\lim \frac{16\theta^2}{\theta}\left(\frac{\sin 4\theta}{4\theta}\right)^2$$.

If so, how do you deduce that $$\lim \left(\frac{\sin 4\theta}{4\theta}\right)^2=1$$ ?

10. Mar 28, 2007

### HallsofIvy

Staff Emeritus
The fact that $$\lim_{\theta\rightarrow 0}\frac{sin \theta}{\theta}= 1$$ together with the fact that x2 is continuous should do it!

11. Mar 28, 2007

### cristo

Staff Emeritus
Indeed. I knew I was missing something pretty obvious!

12. Mar 28, 2007

thank you for your replies, you have been of great help

13. Mar 29, 2007

### jing

$$\lim_{\theta\rightarrow 0} \frac{\sin^24\theta}{\theta}=\lim_{\theta\rightarrow 0} \frac{16\theta^2}{\theta}=\lim_{\theta\rightarrow 0}16\theta=0$$

Last edited: Mar 29, 2007
14. Mar 29, 2007

### VietDao29

How did you go from the first expression to the second expression?
$$\sin \theta \neq \theta$$

15. Mar 29, 2007

### sutupidmath

you need to read the whole threads on this topic explaining how to find this limit. I also explained this on my thread, so read it first, if you still have any questions than after that ask them.

16. Mar 29, 2007

### HallsofIvy

Staff Emeritus
The Taylor's expansion for $sin(\theta)$ is
[tex]\theta- \frac{1}{2}\theta^2+ \frac{1}{3!}\theta^3- \cdot\cdot\cdot[/itex]

The "linear approximation" (tangent line) for $sin(\theta)$ is just $\theta$ so for $sin(4\theta)$ is $4\theta$ and for $sin^2(4\theta)$ is $16\theta^2$.

Since the "higher order" terms will go to 0 faster than the first order term, it is sufficient to use the linear approximate to find the limit. (It is really the same as using L'Hopital's rule.)