limx→−1(x2 − 5x − 2) = 4 Prove by using the epsilon-delta proof.

Hello!

We have to prove that for any e>0 there is d>0 such that |(x^2-5x-2) - 4|<e for all x such that |x-(-1)|<d.

Let's do this: x-(-1)=x+1.

|(x^2-5x-2) - 4| = |x^2-5x-6| = |(x+1)*(x-6)| = |x+1|*|x-6| < d*|x-6| < d*7.

(here we suppose that d<1, so |x-6| < 7)

So for any e>0 we take d = min(1, e/7), which is >0. Then |(x^2-5x-2) - 4| < d*7 <= e.

What is d and how did you got 7 as min?

I use e for epsilon and d for delta.

7 actually shiuld be 8, you are right:|x-6|=|x+1-7| <= |x+1| + 7 < d + 7 <= 8. (if d<=1)

*should

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