Question

Another epsilon-delta problem.

(In answers so I can use equation editor)

Answer 1

Determine the limit l for the given a, and prove that it is the limit by showing how to find a delta such that |f(x)-l|<epsilon for all x satisfying 0<|x-a|<delta.

\[f(x)=x^2+5x-2, a=2\]
I came up with:
\[\delta = {\epsilon \over (x+7)}\]

Answer 2

\[|x^2+5x-2-l|<\epsilon\]
\[|x^2+5x-14|<\epsilon\]
\[|(x+7)(x-2)|<\epsilon\]
\[0<|(x-2)|<\delta \]
\[\delta={\epsilon\over(x+7)}\]
Can anyone corroborate or correct this?

Answer 3

I believe you generally don't want x's in your definition for delta.

Answer 4

That's what I kind of thought, I'm still trying to figure out how this works.

Answer 5

Since we are taking the limit as x goes to 2, we know the value of x will be really close to 2. This will put the value of x+7 really close to 9. So we can say:\[\left|(x+7)(x-2)\right|<\left|10(x-2)\right|<\epsilon\]if \[|x-2|<1\](i should probably switch the order on that).
Then your final delta will be:
\[\delta = \min (\frac{\epsilon}{10},1)\]

Answer 6

Ohh, that totally makes sense. Thanks!