Math Help - delta epsilon

1. delta epsilon

USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT
lim -3x+7=10
x-(-1)

2. Originally Posted by tumi
USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT
lim -3x+7=10
x-0
Find any post by this guy: http://www.mathhelpforum.com/math-he.../krizalid.html

eg. http://www.mathhelpforum.com/math-he...evolution.html

Click on the obvious link in his signature.

3. Doesn't $-3x+7 \rightarrow+7$?

Since $-3x+7$ is a continuous function it's limit is equal to when $x=0$. This gives +7.

4. As Showcase_22 says, you cannot prove $\lim_{x \rightarrow 0} -3x+ 7= 10$- it isn't true!
What you can prove is either $\lim_{x\rightarrow 0} -3x+ 7= 7$ or $\lim_{x\rightarrow -1} -3x+ 7= 10$. Is your problem either of those?

5. Originally Posted by mr fantastic

Click on the obvious link in his signature.
Well in this section that thread is stickied.

6. Yes indeed. There is a typo or something in your statement.

Let's make up something similar and go with that as an example.

$\lim_{x\to 2}[2x-7]=-3$

$|(2x-7)+3|=|2x-4|=2|x-2|<{\epsilon} \;\ if \;\ |x-2|<\frac{\epsilon}{2}, \;\ so \;\ {\delta}=\frac{\epsilon}{2}$

Here's another. I am just making an example up to illustrate to you as a tutorial. Hope it helps.

$\lim_{x\to 4}\sqrt{x}=2$

$|\sqrt{x}-2|=\left|\frac{\sqrt{x}-2}{1}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}\right|=\fr ac{|x-4|}{\sqrt{x}+2}$

Now, if we pull the reins back on delta and restrict it so that ${\delta}\leq 4$, then:

$|x-4|<4$
$0
$0
$2<\sqrt{x}+2<\sqrt{8}+2$
$\frac{1}{2}>\frac{1}{\sqrt{x}+2}>\frac{1}{\sqrt{8} +2}$
$\frac{1}{\sqrt{x}+2}<\frac{1}{2}$
$\frac{|x-4|}{\sqrt{x}+2}\leq \frac{1}{2}|x-4|$

Therefore, thus, and all that:

$|\sqrt{x}-2|<{\epsilon} \;\ if \;\ \frac{1}{2}|x-4|<{\epsilon}, \;\ or \;\ if \;\ |x-4|<2{\epsilon}, \;\ so \;\ {\delta}=min(2{\epsilon},4)$

There.....does that help finish the one you have assuming you fix the typo or what not?.