[tex]\displaystyle\lim_{x\to3}x^3-4x+9=24[/tex]
is to say that for any [tex]\varepsilon>0[/tex], there exists [tex]\delta[/tex] such that
[tex]0<|x-3|<\delta\implies|x^3-4x+9-24|<\varepsilon[/tex]
Notice that
[tex]|x^3-4x-15|=|(x-3)(x^2+3x+5)|=|x-3||x^2+3x+5|=|x-3|\left|\left(x+\dfrac32\right)^2+\dfrac{11}4\right|[/tex]
If we fix [tex]0<\delta\le1[/tex], we would have
[tex]|x-3|<\delta\le1\implies-1<x-3<1\implies2<x<4[/tex]
[tex]\implies15<\left(x+\dfrac32\right)^2+\dfrac{11}4<33[/tex]
which is to say we can place a bound on the quadratic term above of 33 so that
[tex]|x^3-4x-15|=|x-3|\left|\left(x+\dfrac32\right)^2+\dfrac{11}4\right|<33|x-3|[/tex]
This suggests we could set [tex]\delta=\mathrm{min}\left\{1,\dfrac\varepsilon{33}\right\}[/tex].
So if we're given [tex]\varepsilon=0.2[/tex], we would choose [tex]\delta\approx0.0061[/tex], and if [tex]\varepsilon=0.1[/tex], we would choose [tex]\delta\approx0.0030[/tex].
