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Real Analysis proof (simple)

  • Thread starter gaborfk
  • Start date
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1. Homework Statement
Using a delta epsilon method prove:
[tex]\mathop {\lim }\limits_{x \to 1 } x^3+2x^2-3x+4= 4[/tex]


3. The Attempt at a Solution
I got so far as breaking the equation into

[tex]=|x||x+3||x-1|[/tex] now how do I bound it? Also, even more basic question, once I found the bound how do I put the whole thing together as a proof? I do not want you to prove this one, but please, if you can provide a link to a similar problem for me to see how it flows from idea, to figuring out the bounding, to the scratch work of figuring out epsilon, to the actual proof write up. Like the whole flow of things.

I am very familiar with "standard" proofs, but very much lost on the real analysis ones... We have a very bad book with no examples but mainly some ideas, definitions and a ton of homework.


Thank you in advance
 

Answers and Replies

1,631
4
well you need to show that

[tex]\forall\epsilon>0,\exists\delta(\epsilon)>0, \ \ such \ \ \ that \ \ \ whenever \ \ \ \\ \ \

0<|x-1|<\delta => |x^{3}+2x^2-3x+4-4|<\epsilon ????[/tex]

Now from here we go:

[tex]|x^{3}+2x^2-3x+4-4|=|x^{3}+2x^2-3x|=|x(x^{2}+2x-3)|=|x||x+3||x-1|<|x||x+3|\delta<|x+3|2\delta<10\delta=\epsilon=>\delta=\frac{\epsilon}{10}[/tex]

lets see, since x-->1 , it is safe to assume that

[tex]0<x<2=>|x|<2[/tex] also [tex]0<x<2/+3 => 3<x+3<5=>|x+3|<5[/tex]
 
HallsofIvy
Science Advisor
Homework Helper
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In order to be completely "safe" one should say something like [tex]\delta[/itex]= minimum of [itex]\left{\frac{\epsilon}{10}, 1\right}[/tex].
 
1,631
4
In order to be completely "safe" one should say something like [tex]\delta[/itex]= minimum of [itex]\left{\frac{\epsilon}{10}, 1\right}[/tex].

Yes, i forgot to add that!!
 

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