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

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....

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

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

Originally Posted by Danny

Your pictures didn't take. Is this your limit

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

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

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

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

$\displaystyle
\lim_{x \to -2} 2x-2 = -6$

not $\displaystyle -8$.