finding d (delta) for a given E(Epsilon)

**janedoe** · Jul 16th 2009, 01:15 PM

Given lim as x-->-2 of 2x-2 is -8, what is the best choice of d such that |(2x-2)-(-8)| < 0.03 whenever |x+2|<d?

--> I tried to solve |(2x-2)-(-8)| to get it in terms of |x+2| but I can't seem to get it.
I had |2x+6| = 2|x+3| but that doesn't help much.

Also:
b) When the value of x is within 1/20 of 1, how close is the value of f(x) = 4x+5 to 9?

--> I have:
0 < |x-1| < 0.05
|(4x+5)-9| < E

But now I'm stuck....

**Jester** · Jul 16th 2009, 01:20 PM

Originally Posted by janedoe

Given

, what is the best choice of d such that

whenever

?

--> I tried to solve |(2x-2)-(-8)| to get it in terms of |x+2| but I can't seem to get it.
I had |2x+6| = 2|x+3| but that doesn't help much.

Also:
b) When the value of x is within

of 1, how close is the value of

to 9?

--> I have:
0 < |x-1| < 0.05
|(4x+5)-9| < E

But now I'm stuck....

Your pictures didn't take. Is this your limit

$\lim_{x \to -3} 2x-2 = -8$

**janedoe** · Jul 16th 2009, 01:35 PM

Originally Posted by Danny

Your pictures didn't take. Is this your limit

$\lim_{x \to -3} 2x-2 = -8$

No it's actually as x --> -2

I'll repost so you can see the rest...

**Jester** · Jul 16th 2009, 01:52 PM

Originally Posted by janedoe

No it's actually as x --> -2

I'll repost so you can see the rest...

Well then, here's your problem

$<br /> \lim_{x \to -2} 2x-2 = -6$

not $-8$ .

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