USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT

lim -3x+7=10

x-(-1)

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- Feb 7th 2009, 03:16 AMtumidelta epsilon
USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT

lim -3x+7=10

x-(-1) - Feb 7th 2009, 03:23 AMmr fantastic
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. - Feb 7th 2009, 04:52 AMShowcase_22
Doesn't $\displaystyle -3x+7 \rightarrow+7$?

Since $\displaystyle -3x+7$ is a continuous function it's limit is equal to when $\displaystyle x=0$. This gives +7. - Feb 7th 2009, 09:40 AMHallsofIvy
As Showcase_22 says, you cannot prove $\displaystyle \lim_{x \rightarrow 0} -3x+ 7= 10$- it isn't true!

What you can prove is either $\displaystyle \lim_{x\rightarrow 0} -3x+ 7= 7$ or $\displaystyle \lim_{x\rightarrow -1} -3x+ 7= 10$. Is your problem either of those? - Feb 7th 2009, 10:20 AMKrizalid
- Feb 7th 2009, 11:15 AMgalactus
Yes indeed. There is a typo or something in your statement.

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

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

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

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

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

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

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

$\displaystyle |x-4|<4$

$\displaystyle 0<x<8$

$\displaystyle 0<x<\sqrt{8}$

$\displaystyle 2<\sqrt{x}+2<\sqrt{8}+2$

$\displaystyle \frac{1}{2}>\frac{1}{\sqrt{x}+2}>\frac{1}{\sqrt{8} +2}$

$\displaystyle \frac{1}{\sqrt{x}+2}<\frac{1}{2}$

$\displaystyle \frac{|x-4|}{\sqrt{x}+2}\leq \frac{1}{2}|x-4|$

Therefore, thus, and all that:

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

There.....does that help finish the one you have assuming you fix the typo or what not?. - Feb 8th 2009, 02:59 AMtumimy apologies
Sorry i made a mistake. [tex]\lim_{x\rightarrow -1} -3x+ 7= 10 is what i need to prove.can you please help me.

- Feb 8th 2009, 05:58 AMKrizalid
At least read the above posts.