Limit proof

  • Thread starter ciubba
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  • #1
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Homework Statement


Prove [tex]lim_{x->4}\frac{x-4}{\sqrt{x}-2}=4[/tex]

Homework Equations


Epsilon\delta definition

The Attempt at a Solution



I can see that a direct evaluation at 4 leads to an indeterminate form, so:

[tex]\frac{x-4}{\sqrt{x}-2}*\frac{\sqrt{x}+2}{\sqrt{x}+2} \; \mbox{simplifies to} \; \sqrt{x}+2 \; \mbox{when} \; x \ne 4[/tex]

Via epsilon\delta definition,
[tex]|\sqrt{x}+2-4|<\epsilon \; \mbox{whenever} \; |x-4|<\delta[/tex]

Expanding the right side,

[tex]-\delta<x-4<\delta[/tex]
[tex]-\delta+4<x<\delta+4[/tex]
[tex]\sqrt{-\delta+4}<\sqrt{x}<\sqrt{\delta+4} \; \mbox{when} \; \delta \leq 4[/tex]
[tex]\sqrt{-\delta+4}-2<\sqrt{x}-2<\sqrt{\delta+4}-2[/tex]

Thus, [tex]\delta=min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}[/tex]

Given this definition of delta, it is elementary to work backwards towards [tex]|\sqrt{x}+2|<\delta[/tex]

Have I made any errors in my proof?
 

Answers and Replies

  • #2
perplexabot
Gold Member
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Hey cuibba. I can't comment on your technique, but you could have used L'hopital's rule. That way I think is easier.
 
  • #3
pasmith
Homework Helper
1,964
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Homework Statement


Prove [tex]lim_{x->4}\frac{x-4}{\sqrt{x}-2}=4[/tex]

Homework Equations


Epsilon\delta definition

The Attempt at a Solution



I can see that a direct evaluation at 4 leads to an indeterminate form, so:

[tex]\frac{x-4}{\sqrt{x}-2}*\frac{\sqrt{x}+2}{\sqrt{x}+2} \; \mbox{simplifies to} \; \sqrt{x}+2 \; \mbox{when} \; x \ne 4[/tex]

Via epsilon\delta definition,
[tex]|\sqrt{x}+2-4|<\epsilon \; \mbox{whenever} \; |x-4|<\delta[/tex]


Expanding the right side,

[tex]-\delta<x-4<\delta[/tex]
[tex]-\delta+4<x<\delta+4[/tex]
[tex]\sqrt{-\delta+4}<\sqrt{x}<\sqrt{\delta+4} \; \mbox{when} \; \delta \leq 4[/tex]
[tex]\sqrt{-\delta+4}-2<\sqrt{x}-2<\sqrt{\delta+4}-2[/tex]
Thus, [tex]\delta=min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}[/tex]

Given this definition of delta, it is elementary to work backwards towards [tex]|\sqrt{x}+2|<\delta[/tex]

Have I made any errors in my proof?
What is [itex]\delta[/itex] when [itex]\epsilon = 10[/itex]? What about when [itex]\epsilon = 10^{-5}[/itex]?

Your task is to prove:

For every [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that if [itex]0 < |x - 4| < \delta[/itex] then [itex]|\frac{x - 4}{\sqrt{x} - 2} - 4| < \epsilon[/itex].

You need to relate your [itex]\delta[/itex] to the [itex]\epsilon[/itex].
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
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Hey cuibba. I can't comment on your technique, but you could have used L'hopital's rule. That way I think is easier.
He could have used l'Hospital's rule (yes, it really is spelled Hospital) but his way is easier. Alternatively, he could have set ##\sqrt{x} = y## and then evaluated
[tex] \lim_{y \to 2} \frac{y^2 - 4}{y-2},[/tex]
because in this form l'Hospital's rule is much nicer to use.
 
  • #5
65
2
Hey cuibba. I can't comment on your technique, but you could have used L'hopital's rule. That way I think is easier.
Ah, I wish I was allowed to use that technique, but our prof. is adamant on teaching delta ep. definitions first.

What is [itex]\delta[/itex] when [itex]\epsilon = 10[/itex]? What about when [itex]\epsilon = 10^{-5}[/itex]?

Your task is to prove:

For every [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that if [itex]0 < |x - 4| < \delta[/itex] then [itex]|\frac{x - 4}{\sqrt{x} - 2} - 4| < \epsilon[/itex].

You need to relate your [itex]\delta[/itex] to the [itex]\epsilon[/itex].
Oops, I mistyped. I meant to say "Given this definition of delta, it is elementary to work backwards towards [tex]
|\sqrt{x}+2\mathbf{-4}|<\boldsymbol{\epsilon}
[/tex] My main question is whether or not it is correct to say that, for any epsilon, delta is [tex]
min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}[/tex]


He could have used l'Hospital's rule (yes, it really is spelled Hospital) but his way is easier. Alternatively, he could have set ##\sqrt{x} = y## and then evaluated
[tex] \lim_{y \to 2} \frac{y^2 - 4}{y-2},[/tex]
because in this form l'Hospital's rule is much nicer to use.
Ah, I didn't think to do a u-sub. I'll definitely try that next time!
 

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