Epsilon Delta Proof of a Limit

  • Thread starter DarthRoni
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  • #1
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Main Question or Discussion Point

Hey there, I'm new to this forum. Today I thought I would brush up on my calculus.
I would just like to know if my method is correct. Is there an easier way to prove this ?
By the way, it's my first time using LaTeX, so bear with me.

I am trying to prove the following :

[tex]
\lim_{x\rightarrow 10} {x^2} = 100
[/tex]
So, I must find a δ in which the following holds

[tex]
\forall\epsilon>0\ \exists\ \delta>0\ such\ that\ 0<|x - 10| < \delta \implies |x^2 - 100|< \epsilon
[/tex]
I observe the following
[tex]|x-10| = |x + (-10)|[/tex] and by triangle inequality, [tex]|x|+|-10| > |x-10|[/tex] We will also note that [tex]|x| + |-10| = |x| + |10|\implies|x|+|10|>|x-10|[/tex]
Now we can find a δ in terms of ε. By reverse triangle inequality,
[tex]|x^2 - 100|\geq||x^2| - |100||\ and\ ||x^2| - |100||>|x^2| - |100|\ and \ |x^2| - |100| = |x|^2 - |10|^2[/tex][tex]\implies\epsilon > |x|^2 - |10|^2[/tex] Using this we can see that,[tex]\frac{\epsilon+|10|^2}{|x|}>|x|\ as\ well\ as \ \frac{\epsilon-|x|^2}{|10|^2}>|10| \ and\ since\ we\ know:\ |x|+|10| > |x-10|,[/tex]
[tex]\implies \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|} > |x-10|[/tex]
So take [tex]\delta = \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|}[/tex]

Is this correct? I would be really grateful if I got some feedback!
Can we also say the function is continuous [itex]\forall c\in\mathbb R[/itex] in the following way.
[tex]\lim_{x\rightarrow c} {x^2} = c^2[/tex] just take [tex]\delta = \frac{\epsilon+|c|^2}{|x|} + \frac{\epsilon-|x|^2}{|c|}[/tex]
 
Last edited:

Answers and Replies

  • #2
Zondrina
Homework Helper
2,065
136
Hey there, I'm new to this forum. Today I thought I would brush up on my calculus.
I would just like to know if my method is correct. Is there an easier way to prove this ?
By the way, it's my first time using LaTeX, so bear with me.

I am trying to prove the following :

[tex]
\lim_{x\rightarrow 10} {x^2} = 100
[/tex]
So, I must find a δ in which the following holds

[tex]
\forall\epsilon>0\ \exists\ \delta>0\ such\ that\ 0<|x - 10| < \delta \implies |x^2 - 100|< \epsilon
[/tex]
I observe the following
[tex]|x-10| = |x + (-10)|[/tex] and by triangle inequality, [tex]|x|+|-10| > |x-10|[/tex] We will also note that [tex]|x| + |-10| = |x| + |10|\implies|x|+|10|>|x-10|[/tex]
Now we can find a δ in terms of ε. By reverse triangle inequality,
[tex]|x^2 - 100|\geq||x^2| - |100||\ and\ ||x^2| - |100||>|x^2| - |100|\ and \ |x^2| - |100| = |x|^2 - |10|^2[/tex][tex]\implies\epsilon > |x|^2 - |10|^2[/tex] Using this we can see that,[tex]\frac{\epsilon+|10|^2}{|x|}>|x|\ as\ well\ as \ \frac{\epsilon-|x|^2}{|10|^2}>|10| \ and\ since\ we\ know:\ |x|+|10| > |x-10|,[/tex]
[tex]\implies \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|} > |x-10|[/tex]
So take [tex]\delta = \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|}[/tex]

Is this correct? I would be really grateful if I got some feedback!
Can we also say the function is continuous [itex]\forall c\in\mathbb R[/itex] in the following way.
[tex]\lim_{x\rightarrow c} {x^2} = c^2[/tex] just take [tex]\delta = \frac{\epsilon+|c|^2}{|x|} + \frac{\epsilon-|x|^2}{|c|}[/tex]
That seems incredibly overcomplicated.

##|x^2-100| = |x-10||x+10| < δ|x+10|##

Now by the triangle inequality:

##|x+10| = |x - 10 + 20| ≤ |x-10| + 20 < δ + 20##

Now bound delta, what can you conclude?
 
  • #3
31
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Can we just say [itex]\epsilon = \delta|x+10|\implies\frac{\epsilon}{|x+10|}=\delta[/itex]
 
  • #4
Zondrina
Homework Helper
2,065
136
Can we just say [itex]\epsilon = \delta|x+10|\implies\frac{\epsilon}{|x+10|}=\delta[/itex]
Bound delta by the simplest real number you can think of ##δ ≤ 1##.

Now what can you conclude?

##δ|x+10| < δ(δ+20)##
 
  • #5
31
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I see now ! [tex]take\ \delta ≤ 1\implies |x+10|≤ 1+2|10|[/tex] and so since we can say[tex]\epsilon = \delta |x+10| \implies \frac{\epsilon}{1+2|10|} = \delta[/tex]
 
Last edited:
  • #6
Zondrina
Homework Helper
2,065
136
Oh is see ! Suppose [tex]|x-10|<1\implies |x+10| < 1 + 2|10|\ and\ |x^2 - 100| < \delta(\delta + 20)[/tex] So we can take [tex]\epsilon = \delta^2 + 20\delta \implies \epsilon = (1+2|10|)^2 + 20(1+2|10|) \implies \frac{\epsilon}{1+2|10|} - 20 = \delta[/tex]
You could go as far as to say:

##δ(δ+20) ≤ 21δ## because ##δ ≤ 1##.

Then you could conclude:
##21δ ≤ ε##
##δ ≤ \frac{ε}{21}##

So ##δ = min\{1, \frac{ε}{21}\}##
 
  • #7
31
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Thanks so your help I realized my answer was wrong after I posted it, Why is it we cannot use x to represent out epsilon ?
 
  • #8
Zondrina
Homework Helper
2,065
136
Thanks so your help I realized my answer was wrong after I posted it, Why is it we cannot use x to represent out epsilon ?
##x## has nothing to do with ##ε##. The objective is to get a ##δ(ε)##.
 
  • #9
31
0
I think I understand now, we are trying to find a bound on [itex]|x-c|[/itex] and it doesn't make sense to include [itex]x[/itex] in the value for [itex]\delta[/itex]
 

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