How do I find the delta from the limit with an epsilon of o.1?

Question

I am  trying to find a delta that corresponds to an epsilon of 0.1.

limx->2 1/x^2 = 1/4

Answer 1

^{Let y=1/x² and x changes to x+∆x, where ∆x is a small change in x.}

^{At the limit x=2, y=¼. What we are going to find is the small change in y (∆y) corresponding to the small change in x. You have used delta and epsilon as the small changes, and I’m going to assume that delta is the same as ∆x and epsilon is the same as ∆y.}

^{So we have:}

^{y=1/x² and y+∆y=1/(x+∆x)².}

^{We find ∆y by subtracting these two equations:}

^{(y+∆y)-y=1/(x+∆x)²-1/x²,}

^{∆y=(x²-(x+∆x)²)/(x²(x+∆x)²),}

^{∆y=(x+x+∆x)(x-x-∆x)/(x²(x+∆x)²),}

^{∆y=-∆x(2x+∆x)/(x²(x+∆x)²).}

^{We can start plugging in values: ∆x=0.1, x=2:}

^{∆y=-0.1(4.1)/(4(2.1)²)=-0.41/17.64=-0.023 approx.}

^{So a change of 0.1 in x when x is close to 2 gives us a change of about -0.023 in 1/x².}

^{One simplification we can use is to reduce the equation for ∆y by ignoring ∆x in x+∆x and 2x+∆x:}

^{∆y≈-2x∆x/x⁴=-2∆x/x³.}

^{So ∆y=-0.2/8=-0.025.}

^{There’s not much difference between -0.025 and -0.023. So the answer is epsilon=-0.025 approximately.}