prove limit by definition

**drugcpp** · November 5th 2010, 03:42 AM

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.

**Prove It** · November 5th 2010, 04:06 AM

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 $\displaystyle 0 < |x - 2| < \delta \implies |x^2 - 4 - 0| < \epsilon$ .

**drugcpp** · November 5th 2010, 05:02 AM

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.

**Prove It** · November 5th 2010, 05:18 AM

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 $\displaystyle \left|\frac{2x+3}{x+2} - \frac{3}{2}\right|<\epsilon$ .

**drugcpp** · November 5th 2010, 05:33 AM

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!

Thread: prove limit by definition

LinkBack

Thread Tools

Display

prove limit by definition

Similar Math Help Forum Discussions

prove limit by definition

prove by limit definition

use limit definition to prove that the limit is correct

Use definition of limit to prove lim (sqrt(1 + n^(-1))) = 1?

1. find the limit 2. use the definition of the limit to prove that

Search Tags