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.