prove limit by definition

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November 5th 2010, 03:42 AM
drugcpp

prove limit by definition

Hello everyone, this is my first topic on this forum, so if by chance this question was asked in the past, redirect me to the relevant post.

So, I understand the definition of function limit, I just can't figure how to solve those problems.
For example, I have
$lim(x->2)(x²-4)=5$

So I write |x²-9|<ε, 0<|x-2|<δ
then
|x-3||x+3|<ε
and what's now? I know that probably I should assume that δ<1 or something, and then I stuck.
If there a place when I can see just a big amount of examples, this would be great.

Thank you.
November 5th 2010, 04:06 AM
Prove It

First of all, $\displaystyle \lim_{x \to 2}(x^2 - 4) = 0$ , not $\displaystyle 5$ .

Now, to prove $\displaystyle \lim_{x \to 2}(x^2 - 4) = 0$ , you need to show $\displaystyle 0 < |x - 2| < \delta \implies |x^2 - 4 - 0| < \epsilon$ .

So now we solve $\displaystyle |x^2 - 4| < \epsilon$ for $\displaystyle |x - 2|$ ...

$\displaystyle |x^2 - 4| < \epsilon$

$\displaystyle |x - 2||x + 2| < \epsilon$

$\displaystyle M|x - 2| < \epsilon$

$\displaystyle |x - 2| < \frac{\epsilon}{M}$ .

We need to choose an $\displaystyle M$ such that $\displaystyle |x + 2| \leq M$ for $\displaystyle x$ close to $\displaystyle 2$ .

If we require $\displaystyle 0 < |x - 2| < 1$ , then

$\displaystyle x - 2 < 1$

$\displaystyle x < 3$

$\displaystyle x + 2 < 5$

$\displaystyle |x + 2| < 5$ .

So we choose $\displaystyle M = 5$ , and thus we define $\displaystyle \delta = \min \left \{ 1, \frac{\epsilon}{5}\right \}$ .

Proof: Let $\displaystyle \epsilon > 0$ and define $\displaystyle \delta = \min \left\{1, \frac{\epsilon}{5} \right\}$ .

Then if $\displaystyle 0 < |x - 2| < \delta$ we have

$\displaystyle |x^2 - 4| = |x - 2||x + 2|$

$\displaystyle < 5|x - 2|$ since $\displaystyle |x - 2| < \delta$ and $\displaystyle \delta \leq 1$

$\displaystyle < 5\left(\frac{\epsilon}{5}\right)$ since $\displaystyle |x - 2| < \delta$ and $\displaystyle \delta \leq \frac{\epsilon}{5}$

$\displaystyle = \epsilon$ .

Thus $\displaystyle 0 < |x - 2| < \delta \implies |x^2 - 4 - 0| < \epsilon$ .
November 5th 2010, 05:02 AM
drugcpp

Thanks! Now to check that I got it, I'll do another example, and you say if I'm right, OK?

lim(x->0) (2x+3)/(x+2) = 3/2

|x/(2x+4)|<ε

we require that 0 < x < 1

so M * x < ε

x < ε/M
and M > 1/(2x+4)

-1 < x < 1
-2 < 2x
2 < 2x+4
1/2 > 1/(2x+4)
M = 1/2

δ = min{1, 2ε}

Right?

BTW, I didn't find how to use Latex in order to write beautiful math expressions. The link in FAQ is broken.
November 5th 2010, 05:18 AM
Prove It

What you have done is fine.

So now you need to do the proof, i.e. now show that if $\displaystyle 0 < |x| < \delta$ then you can get $\displaystyle \left|\frac{2x+3}{x+2} - \frac{3}{2}\right|<\epsilon$ .
November 5th 2010, 05:33 AM
drugcpp

In my course I'm not required to provide the proof part. Maybe it's because we have a lot of material to learn and a small amount of time. I'm glad that I did it right, thank you for your explanation!