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Delta epsilon limits

  • Thread starter Shelby
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Homework Statement



Given that lim f(x)=L as x approaches a , prove that lim (x+f(x))=a+L as x approaches aView attachment 9630. Your proof cannot assume that the limit of a sum of two functions is the sum of their individual limits. You must use the delta-epsilon definition of limit in your proof.
and
Given that lim f(x)=L as x approaches a , prove that lim x*f(x)=aL as x approaches a

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The Attempt at a Solution


attempt is in the attatchment
 

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Answers and Replies

  • #2
VietDao29
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Next, the negative must be distributed to the “a” and to the “L” of (a + L) and an L must be substituted for f(x) because it is given that the [tex]\lim_{x \rightarrow a} f(x) = L[/tex]
|x + L - a - L|...
That part is wrong, you cannot sub L for f(x) like that, it's not correct.

Now, back at the start, the problem gives that:
[tex]\lim_{x \rightarrow a} f(x) = L[/tex], and you must use this to prove: [tex]\lim_{x \rightarrow a} [x + f(x) ] = a + L[/tex]
[tex]\lim_{x \rightarrow a} f(x) = L[/tex] means that, for any arbitrary small [tex]\epsilon[/tex], there exists a [tex]\delta[/tex], such that:
[tex]0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon[/tex]


What we should prove is:
For any arbitrary small [tex]\epsilon_1[/tex], there exists a [tex]\delta_1[/tex], such that:
[tex]0 < |x - a| < \delta_1 \Rightarrow |[x + f(x)] - [a + L]| < \epsilon_1[/tex]

We will deal with this part:
[tex]|[x + f(x)] - [a + L]| = |(x - a) + [f(x) - L]| < |x - a| + |f(x) - L|[/tex]


Now we choose, [tex]\epsilon = \frac{\epsilon_1}{2}[/tex].

Now, there should exists a [tex]\delta[/tex], such that: [tex]0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon = \frac{\epsilon_1}{2}[/tex]
Choose [tex]\delta_1[/tex], so that: [tex]\delta_1 = \mbox{min} \left( \delta, \frac{\epsilon_1}{2} \right)[/tex]

[tex]\delta_1 = \mbox{min} \left( \delta, \frac{\epsilon_1}{2} \right) \Rightarrow \delta_1 \leq \delta[/tex], so we have:
[tex]0 < |x - a| < \delta_1 \leq \delta \Rightarrow |f(x) - L| < \epsilon = \frac{\epsilon_1}{2}[/tex]

[tex]\delta_1 = \mbox{min} \left( \delta, \frac{\epsilon_1}{2} \right) \Rightarrow \delta_1 \leq \frac{\epsilon}{2}[/tex]


Now, if we have:
[tex]0 < |x - a| < \delta_1 \Rightarrow |[x + f(x)] - [a + L]| < |x - a| + |f(x) - L| < \delta_1 + \epsilon < \frac{\epsilon_1}{2} + \frac{\epsilon_1}{2} = \epsilon_1[/tex], so:
[tex]\lim_{x \rightarrow a} [x + f(x) ] = a + L[/tex] (Q.E.D)

Is there any where unclear?
Can you do the same to part II? :)
 
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  • #3
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Thank you, I do have one question

Why did you choose to divide epsilon 1 by 2?
 
  • #4
mjsd
Homework Helper
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because eventually you wanna sum up two of them.. and 1/2 +1/2 =1 gives you just one epsilon... and this matches the target you wanna reach. that's all
 
  • #5
VietDao29
Homework Helper
1,423
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Why did you choose to divide epsilon 1 by 2?
Since you want to prove that:
[tex]|[x + f(x)] - [a + L]| = |(x - a) + [f(x) - L]| < |x - a| + |f(x) - L| < \epsilon_1[/tex], i.e, the sum of |x - a|, and |f(x) - L|, so it's natural to choose the sum of [tex]\frac{\epsilon_1}{2}[/tex], and [tex]\frac{\epsilon_1}{2}[/tex], as [tex]\frac{\epsilon_1}{2} + \frac{\epsilon_1}{2} = \epsilon_1[/tex].
Then assign [tex]\delta, \ \delta_1 , \ \epsilon[/tex] wisely, so that: [tex]|x - a| < \frac{\epsilon_1}{2}[/tex], and [tex]|f(x) - L| < \frac{\epsilon_1}{2}[/tex]
 
  • #6
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I think I understand the process shown for the above proof but if someone could provide the second part it would improve my understanding. Given that lim f(x)=L as x approaches a , prove that lim x*f(x)=aL as x approaches a using the delta epsilon definition of a limit. This would be much appreciated. Thanks.
 
  • #7
169
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just try it yourselves! Use the method and I'm confident it will work out!
 

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