: Limits

  • #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
 
Last edited:

Answers and Replies

  • #2
Gib Z
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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
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As far as i can see part A has no question to solve, it simply states the definition of the limit
 
  • #4
Gib Z
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I think it is asking to prove that statement it provided or something
 
  • #5
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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
Gib Z
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Well my graphing program says the limit is zero..
 
  • #8
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4
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 ([email protected])/([email protected]), what do you get? then write the numerator as (4*@)^2, so you will get (sin([email protected])/[email protected])^2, and the limit of this is 1, but we have still the limit of [email protected] as @->0 so the result is 0.
sorry for my symbols, i hope u understand @ stands for theta, the angle at sin
 
  • #9
cristo
Staff Emeritus
Science Advisor
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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 ([email protected])/([email protected]), what do you get? then write the numerator as (4*@)^2, so you will get (sin([email protected])/[email protected])^2, and the limit of this is 1, but we have still the limit of [email protected] 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
HallsofIvy
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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
cristo
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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
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[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]
 
Last edited:
  • #14
VietDao29
Homework Helper
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[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
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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
HallsofIvy
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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.)
 

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