Given ε>0, we need to find δ(ε) such that |1/(x+3)−1/5| < ε for |x−2| < δ
i.e. |0.2*(2−x)/(x+3)| < ε for |x−2| < δ … (i)
The standard approach is to find an inequality, independent of x, that must be satisfied by δ and then compare with (i) and pick suitable value for δ.
Any suitable δ implies |0.2*(2−x)/(x+3)| < 0.2δ/|x+3|
The RHS of this can be made independent of x by finding a lower bound for |x+3|. This is done by restricting x to |x−2|<1 and using the triangle inequality. We can do this since we are only interested in the behaviour near x=2.
|x+3| + |2−x| ≥ 5 → |x+3| ≥ 5 − |2−x| → |x+3| ≥ 4
So |0.2*(2−x)/(x+3)| < 0.2δ/4 = δ/20
Comparing this with (i) we see that if we put δ=20ε then (i) is satisfied.
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You want to show that lim [x→2] 1/(x+3) = 1/5
Given ε>0, we need to find δ(ε) such that |1/(x+3)−1/5| < ε for |x−2| < δ
i.e. |0.2*(2−x)/(x+3)| < ε for |x−2| < δ … (i)
The standard approach is to find an inequality, independent of x, that must be satisfied by δ and then compare with (i) and pick suitable value for δ.
Any suitable δ implies |0.2*(2−x)/(x+3)| < 0.2δ/|x+3|
The RHS of this can be made independent of x by finding a lower bound for |x+3|. This is done by restricting x to |x−2|<1 and using the triangle inequality. We can do this since we are only interested in the behaviour near x=2.
|x+3| + |2−x| ≥ 5 → |x+3| ≥ 5 − |2−x| → |x+3| ≥ 4
So |0.2*(2−x)/(x+3)| < 0.2δ/4 = δ/20
Comparing this with (i) we see that if we put δ=20ε then (i) is satisfied.
Adding the restriction |x−2|<1 gives δ=min(1,20ε)