Question

For the limit lim x → 3 (x3 − 4x + 9) = 24 illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1. (round your answers to four decimal places.)

Answer 1

$\displaystyle\lim_{x\to3}x^3-4x+9=24$

is to say that for any $\varepsilon>0$, there exists $\delta$ such that

$0<|x-3|<\delta\implies|x^3-4x+9-24|<\varepsilon$

Notice that

$|x^3-4x-15|=|(x-3)(x^2+3x+5)|=|x-3||x^2+3x+5|=|x-3|\left|\left(x+\dfrac32\right)^2+\dfrac{11}4\right|$

If we fix $0<\delta\le1$, we would have

$|x-3|<\delta\le1\implies-1<x-3<1\implies2<x<4$

$\implies15<\left(x+\dfrac32\right)^2+\dfrac{11}4<33$

which is to say we can place a bound on the quadratic term above of 33 so that

$|x^3-4x-15|=|x-3|\left|\left(x+\dfrac32\right)^2+\dfrac{11}4\right|<33|x-3|$

This suggests we could set $\delta=\mathrm{min}\left\{1,\dfrac\varepsilon{33}\right\}$.

So if we're given $\varepsilon=0.2$, we would choose $\delta\approx0.0061$, and if $\varepsilon=0.1$, we would choose $\delta\approx0.0030$.