How Does the Definition of a Limit Apply to Functions and Trigonometric Limits?

[unique_pavadrin]

URGENT: Limits

Homework Statement

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 &lt; \left| {x - c} \right| &lt; \delta <br /> 0 &lt; \left| {f\left( x \right) - L} \right| &lt; \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):
<br /> \begin{array}{l}<br /> \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta } = \frac{{\sin ^2 4\left( 0 \right)}}{0} = \frac{0}{0} \\ <br /> \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&#039;\left( \theta \right) = 1 \\ <br /> f\left( \theta \right) = \sin ^2 4\theta \\ <br /> \sin ^2 \theta = \frac{{1 - \cos 2\theta }}{2} \\ <br /> f\left( \theta \right)\sin ^2 4\theta = \frac{{1 - \cos 8\theta }}{2} \\ <br /> \end{array}<br />

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

 

[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.

 

[FunkyDwarf]

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

 

[Gib Z]

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

 

[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

 

[Gib Z]

Well my graphing program says the limit is zero..

 

[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, don't know how to use the mathematical notation thingy

 

[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

 

[cristo]

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

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 ?

 

[HallsofIvy]

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

 

[cristo]

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

Indeed. I knew I was missing something pretty obvious!

 

[unique_pavadrin]

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

 

[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

 

[VietDao29]

\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

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

 

[sutupidmath]

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

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.

 

[HallsofIvy]

The Taylor's expansion for sin(\theta) is
\theta- \frac{1}{2}\theta^2+ \frac{1}{3!}\theta^3- \cdot\cdot\cdot[/itex]<br /> <br /> The &quot;linear approximation&quot; (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.<br /> <br /> Since the &quot;higher order&quot; 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&#039;Hopital&#039;s rule.)