Question

For the limit

lim
x → 2
(x3 − 5x + 5) = 3
illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1.

Answer 1

`\displaystyle\lim_(x\to2)(x^3-5x+5)=3`

means there exists some
`\delta` such that whenever
`0<|x-\delta|<2`, it's guaranteed that
`|x^3-5x+5-3|<\epsilon` for all
`\epsilon>0`.

You have


`|x^3-5x+5-3|=|x^3-5x+2|=|(x-2)(x^2+2x-1)|=|x-2||x^2+2x-1|`

Completing the square for the quadratic term yields


`|x-2||(x+1)^2-2|`

and by the triangle inequality,


`|x-2||(x+1)^2-2|\le|x-2||x+1|^1+2|x-2|=|x-2|(|x+1|^2+2)`

Suppose we fix
`\delta\le1`. Then if
`|x-2|<\delta\le1`, it follows that
`|x+1|\le4`, since


`|x-2|\le1\implies-1\le x-2\le1\implies 2\le x+1\le4`

This means


`|x-2||(x+1)^2-2|\le|x-2|(|x+1|^2+2)\le|x-2|(4^2+2)=18|x-2|<\epsilon`

`\implies|x-2|<\delta=\frac\epsilon{18}`

So to guarantee that
`x^3-5x+5` is within
`\epsilon` of the limit, we can choose the smaller of the two choices for
`\delta`, or
`\delta=\min\left\{1,\frac\epsilon{18}\right\}`.

Now, when
`\epsilon=0.2=\frac15`, you would have
`\delta=\min\left\{1,\frac1{90}\right\}=\frac1{90}\approx0.0111`, and when
`\epsilon=0.1=\frac1{10}`, you would have
`\delta=\min\left\{1,\frac1{180}\right\}=\frac1{180}\approx0.0056`.