prove limit by definition

• Nov 5th 2010, 04:42 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.
• Nov 5th 2010, 05:06 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 $\displaystyle 0 < |x - 2| < \delta \implies |x^2 - 4 - 0| < \epsilon$.
• Nov 5th 2010, 06:02 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.
• Nov 5th 2010, 06:18 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 $\displaystyle \left|\frac{2x+3}{x+2} - \frac{3}{2}\right|<\epsilon$.
• Nov 5th 2010, 06:33 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!