Limits on f(x) and \theta

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In summary, part A asks for a solution to f(x) being a function defined on an open interval containing c. 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 For part B, the working is as follows: f(x) = sin^2 4x, where sin^2(x) is the second order derivative of sin(x). The limit for sin(\theta) as \theta \right
  • #1
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:

[tex]\mathop {\lim }\limits_{x \to a} f\left( x \right) = L[/tex]

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

PART B
Hence, evaluate exactly [tex]\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta }[/tex]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):
[tex]
\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}
[/tex]

Am i doing this part correctly? Is there an easier way? Many thanks to all help and suggestions offered.
unique_pavadrin
 
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  • #2
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
As far as i can see part A has no question to solve, it simply states the definition of the limit
 
  • #4
I think it is asking to prove that statement it provided or something
 
  • #5
Is it not possible to use small angles [tex] {\sin \theta }\approx\theta[/tex]

so [tex] \frac{{\sin ^2 4\theta }}{\theta }\approx16\theta[/tex] as [tex]\theta \rightarrow 0[/tex]
 
  • #6
Well my graphing program says the limit is zero..
 
  • #7
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
 
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  • #8
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
sutupidmath said:
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? [tex]\lim_{x\rightarrow 0} \frac{\sin^24\theta}{\theta}=\lim \frac{16\theta^2}{\theta}\left(\frac{\sin 4\theta}{4\theta}\right)^2[/tex].

If so, how do you deduce that [tex]\lim \left(\frac{\sin 4\theta}{4\theta}\right)^2=1[/tex] ?
 
  • #10
The fact that [tex]\lim_{\theta\rightarrow 0}\frac{sin \theta}{\theta}= 1[/tex] together with the fact that x2 is continuous should do it!
 
  • #11
HallsofIvy said:
The fact that [tex]\lim_{\theta\rightarrow 0}\frac{sin \theta}{\theta}= 1[/tex] together with the fact that x2 is continuous should do it!

Indeed. I knew I was missing something pretty obvious!
 
  • #12
thank you for your replies, you have been of great help
unique_pavadrin
 
  • #13
[tex]\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[/tex]
 
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  • #14
jing said:
[tex]\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[/tex]
How did you go from the first expression to the second expression? :bugeye:
[tex]\sin \theta \neq \theta[/tex]
 
  • #15
VietDao29 said:
How did you go from the first expression to the second expression? :bugeye:
[tex]\sin \theta \neq \theta[/tex]

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

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

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.)
 

What is the definition of a limit?

The limit of a function f(x) as x approaches a value c is the value that the function gets closer and closer to as x gets closer and closer to c. It can be thought of as the "end behavior" of the function at that particular point.

How do you find the limit of a function?

To find the limit of a function, you can plug in values of x that are closer and closer to the given value c and observe the corresponding values of the function. If the values of the function are getting closer and closer to a single value, that value is the limit.

What is the significance of limits in calculus?

Limits play a crucial role in calculus as they allow us to study the behavior of a function at a particular point. They also allow us to define derivatives and integrals, which are fundamental concepts in calculus.

Can a limit of a function exist if the function is not defined at that point?

Yes, a limit can exist even if the function is not defined at that point. This is because the limit only considers the behavior of the function as x approaches a particular value and does not depend on the actual value of the function at that point.

How can you determine if a limit exists or not?

A limit exists if the values of the function approach a single value as x gets closer and closer to the given value c. If the values of the function approach different values from the left and right sides of c, then the limit does not exist. Additionally, if the values of the function become infinitely large or approach different values as x approaches c, then the limit also does not exist.

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