USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT

lim -3x+7=10

x-(-1)

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

lim -3x+7=10

x-(-1) - February 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. - February 7th 2009, 04:52 AMShowcase_22
Doesn't ?

Since is a continuous function it's limit is equal to when . This gives +7. - February 7th 2009, 09:40 AMHallsofIvy
As Showcase_22 says, you cannot prove - it isn't true!

What you can prove is either or . Is your problem either of those? - February 7th 2009, 10:20 AMKrizalid
- February 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.

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

Now, if we pull the reins back on delta and restrict it so that , then:

Therefore, thus, and all that:

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

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