Prove A Left Hand Limit

  • Thread starter TrueStar
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  • #1
95
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Homework Statement



Proove the statement using the [tex]\epsilon[/tex] and [tex]\delta[/tex] definition of a limit.

The limit of x as it approaches 9 from the left side of the x axis is [tex]\sqrt[4]{9-x}[/tex]=0


Homework Equations



The limit of x as it approaches a from the left side of the x axis is f(x) = L

if for every number [tex]\epsilon[/tex]>0 there is a number [tex]\delta[/tex]>0 such that if

a-[tex]\delta[/tex]<x<a then |f(x)-L|<[tex]\epsilon[/tex]

The Attempt at a Solution



9-[tex]\delta[/tex]<x<9 and |[tex]\sqrt[4]{9-x}[/tex]-0|<[tex]\epsilon[/tex]

Using algebra, I've worked my way to |9-x|<[tex]\epsilon^{4}[/tex]. I'm not sure where to go with it now. Do I subtract the 9 over and then change the sign of x, ending up with |x|>-[tex]\epsilon[/tex][tex]^{4}[/tex]+9 ? That doesn't seem right to me.
 

Answers and Replies

  • #2
You can pick delta=episilon^4 and you are done.
 
  • #3
95
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That's great! However, I don't know why it's OK to stop there. Would anyone be kind enough to explain?

More specifically, why can I stop at |9-x| and how do I work my way back to verify that [tex]\epsilon[/tex][tex]^{4}[/tex] can be substituted for [tex]\delta[/tex]?
 
  • #4
You can stop there because this is the precise definitionof a limit. We say limit as x approaches a f(x)=L if
When|f(x)-L|< episilon then |x-a|< delta.

If your case a=7 and delta = episilon^4.

Too lazy to use latex of my phone, sorry!
 
  • #5
I believe it's because if we assume that d = E^4, then we have 9 - E^4 < x < 9. This means that 9 - E^4 - x < 0 < 9 - x. Then 9 - x < E^4 < 9 - x + E^4.

Then you have the algebra: since |(9 - x)^(1/4)| < E, then |9 - x| < E^4 in order for the limit to be true. We've found a delta such that this is true, so our proof is complete.
 
  • #6
95
0
Thank you both so much for explaining it for me! I understand why it works now. There were no good examples in the textbook, so I can use this as a guide for similar problems.
 

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