Limit question

  • #1
I made the following Limit problem by myself and just for practice. However, I found some question on it. Please take a look:

[tex]lim_{x\rightarrow 2} (x^2 - 9) = -5[/tex]

And the solutions (by myself):

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

[tex]|(x^2 - 9) - (-5)| < \epsilon[/tex]

[tex]|x^2 - 9 + 5| < \epsilon[/tex]

[tex]|x^2 - 4| < \epsilon[/tex]

[tex]-\epsilon < x^2 - 4 < \epsilon[/tex]

[tex]-\epsilon + 4 < x^2 < \epsilon + 4[/tex]

It gets more complicated since the inequality in the QUADRATIC form.

Thus,

[tex]\sqrt{-\epsilon + 4} < \sqrt{x^2} < \sqrt{\epsilon + 4}[/tex]

[tex]\sqrt{-\epsilon + 4} < |x| < \sqrt{\epsilon + 4}[/tex]

Divide by 2 form:

FIRST

[tex]-\sqrt{\epsilon + 4} < x < \sqrt{\epsilon + 4}[/tex]

SECOND

[tex] x > \sqrt{-\epsilon + 4}[/tex] and [tex]x < -\sqrt{-\epsilon + 4}[/tex]

So, can anyone check my answer whether it’s true or not?
 

Answers and Replies

  • #2
mathwonk
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since limits respect addition, it sufficies to show lim x^2 = 4, as x-->2.

but if we write x = 2+h, then x^2 = 4+2h + h^2.

then if we choose |h| both less than 1 (so that squaring it makes it smaller), and also less than epsilon/4, then |2h + h^2| is le 2|h| +|h| is le 3epsilon/4 < epsilon.


this is for me an easier way to do these problems. i.e. estimate f(a+h) - f(a), and show it goes to zero.
 
  • #3
HallsofIvy
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I made the following Limit problem by myself and just for practice. However, I found some question on it. Please take a look:

[tex]lim_{x\rightarrow 2} (x^2 - 9) = -5[/tex]

And the solutions (by myself):

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

[tex]|(x^2 - 9) - (-5)| < \epsilon[/tex]

[tex]|x^2 - 9 + 5| < \epsilon[/tex]

[tex]|x^2 - 4| < \epsilon[/tex]

[tex]-\epsilon < x^2 - 4 < \epsilon[/tex]
Great so far!

[tex]-\epsilon + 4 < x^2 < \epsilon + 4[/tex]

It gets more complicated since the inequality in the QUADRATIC form.
Right- so don't do that!!

Instead write it as
[tex]-\epsilon< (x-2)(x+2)<\epsilon[/tex]
Now, obviously what you want is something like
[tex]-\delta< x- 2< \delta[/tex]
so the question now is what to do with that "x+2".
If you could guarentee, for example, that 1< x+2< 2, then you could say
[tex]1(-\delta< (x-2)(x+2)< 2\delta[/tex]
and that will give
[tex]-\epsilon< (x-2)(x+2)<\epsilon[/tex]
as long as [itex]\delta< (1/2)\epsilon[/itex].

To make sure that 1< x+2< 2, you must have 1-4< x+2-4< 2-4 or -3< x-2< -2 which will be true if |x-2|< 2. Choose [itex]\delta[/itex] to be the smaller of 2 or [itex](1/2)\epsilon[/itex] to make sure both inequalities are true.


Thus,

[tex]\sqrt{-\epsilon + 4} < \sqrt{x^2} < \sqrt{\epsilon + 4}[/tex]

[tex]\sqrt{-\epsilon + 4} < |x| < \sqrt{\epsilon + 4}[/tex]

Divide by 2 form:

FIRST

[tex]-\sqrt{\epsilon + 4} < x < \sqrt{\epsilon + 4}[/tex]

SECOND

[tex] x > \sqrt{-\epsilon + 4}[/tex] and [tex]x < -\sqrt{-\epsilon + 4}[/tex]

So, can anyone check my answer whether it’s true or not?
 

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