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Proving Limit Laws

  • Thread starter ciubba
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  • #1
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Homework Statement


Prove that [tex]\lim_{x->a}[f(x)+g(x)]=\lim_{x->a}[f(x)]+\lim_{x->a}[g(x)][/tex]

Homework Equations


Epsilon/delta definition

The Attempt at a Solution



The book says:

Let the limit of f(x)=L and the limit of g(x)=M. Then,

[tex]\mid f(x)-L \mid<\frac{\epsilon}{2}[/tex] whenever [tex]0<\mid x-a \mid<\delta_{1}[/tex]
and
[tex]\mid g(x)-M \mid<\frac{\epsilon}{2}[/tex] whenever [tex]0<\mid x-a \mid<\delta_{2}[/tex]

Let [tex] \delta=min \big\{ \delta_{1} ,\delta_{2} \big\}[/tex] and suppose [tex]0<\mid x-a \mid<\delta[/tex]. Because [tex]\delta\leq\delta_{1}[/tex], it follows that [tex]0<|x-a|\delta_{1}[/tex] and [tex]|f(x)-L<\frac{\epsilon}{2}[/tex].

Similarly, [tex]\delta\leq\delta_{2}[/tex], it follows that [tex]0<|x-a|\delta_{2}[/tex] and [tex]|g(x)-L<\frac{\epsilon}{2}[/tex]. Therefore,

[tex]|[f(x)+g(x)]-(L+M)|<\epsilon[/tex], which implies that [tex]]\lim_{x->a}[f(x)+g(x)]=L+M=\lim_{x->a}[f(x)]+\lim_{x->a}[g(x)][/tex]

My biggest question is from where they obtained a value of epsilon/2 from. I feel that I cannot even begin to understand the proof until I understand that point.

Edit: The latex didn't come out as clearly as I'd hoped, so here is a picture: http://s9.postimg.org/7gfz3owul/Proof.gif
 
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Answers and Replies

  • #2
ehild
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Read the first sentence after SOLUTION in the picture: "Assume that ε > 0 is given " You can require that both functions deviate from their limit less than half of the given ε.
 
  • #3
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Read the first sentence after SOLUTION in the picture: "Assume that ε > 0 is given " You can require that both functions deviate from their limit less than half of the given ε.
Yes, I saw that, but I also could have defined it as < ε/3 or 2piε/4. What is the significance of making it ε/2? I looked at a proof for the product limit laws and saw a similar technique employed.
 
  • #4
ehild
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You have to find δ so as the deviation of the combined function f(x)+g(x) from its limit is less than ε. The simplest is to choose equal ε1 and ε2 for the functions f and g, so as ε1+ε2 = ε. But you have the freedom to choose ε/3 for one function and 2/3 ε for the other one, or anything else if they are positive and their sum is ε.
 
  • #5
Ray Vickson
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Yes, I saw that, but I also could have defined it as < ε/3 or 2piε/4. What is the significance of making it ε/2? I looked at a proof for the product limit laws and saw a similar technique employed.
It really does not matter. For any (small) ##\epsilon' > 0## we can find ##\delta > 0## such that ##|f(x)-L| < 3 \epsilon'## and ##|g(x) -M |< 2 \pi \epsilon'## for ##|x - a| < \delta##. Then we have
[tex] |f(x) + g(x) - (L+M)| < (3 + 2 \pi) \epsilon'[/tex]
for ##|x-a| < \delta##. Now just put ##\epsilon = (3 + 2 \pi) \epsilon'##. Given any (small) ##\epsilon > 0## we can thus find ##\delta > 0## such that ##|f(x) + g(x) - (L+M)| < \epsilon## if ##|x-a| < \delta##. We do that by putting ##\epsilon' = \epsilon/(3 + 2 \pi)## and using the ##\delta## that goes along with ##\epsilon'##. It is just a lot neater to put ##\epsilon' = \epsilon/2##.
 
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