Given lim_(x->2) (2x-2) = -6, what is the best choice of d such that |(2x-2)-(-6)| < 0.03 whenever |x+2| < d?
I think you mean lim_(x -> -2)
You want
|(2x-2)-(-6)| < 0.03
|2x+4| < 0.03
|x+2| < 0.015
Yep.
I've gone to the point -2.015<x<-1.985, because -0.03<2x+4<0.03 and you solve from that point on. But what do I choose to get d?
well, duh. d = 0.015
look at what I wrote.
To find the best choice of d, we need to analyze the expression |(2x-2)-(-6)| < 0.03 when |x+2| < d.
First, let's simplify the expression inside the absolute value:
|(2x-2)-(-6)| = |2x + 4|
Now, let's rewrite the inequality using the simplified expression:
|2x + 4| < 0.03
Since the value inside the absolute value must be less than 0.03, we can split it into two cases: positive and negative.
1. Positive case: 2x + 4 < 0.03
Simplifying this inequality, we get: 2x < -3.97
Solving for x, we find: x < -1.985
2. Negative case: -(2x + 4) < 0.03
Simplifying this inequality, we get: -2x - 4 < 0.03
Solving for x, we find: x > -2.015
Now, let's find the range of x values that satisfy both cases:
-1.985 < x < -2.015
To find the best choice of d, we need to determine the maximum distance between the critical points (-1.985 and -2.015) and the given point of interest x = 2.
The maximum distance will be half the difference between the two critical points:
d = (|-1.985 - 2.015|) / 2
d = 0.03 / 2
d = 0.015
Therefore, the best choice of d is 0.015, such that |(2x-2)-(-6)| < 0.03 whenever |x+2| < d.