So I want to show that $\lim_{x \to 8} \dfrac{3x^{1/3}}{x+4} = \dfrac{1}{2}$ using an $\epsilon-\delta$ proof. So far, this is what I've done.
By the $\epsilon-\delta$ definition of a limit, $\lim_{x\to a}f(x) =L \iff \forall \epsilon >0, \exists \delta>0 \space (0 < |x-a| < \delta \space\wedge |f(x)-L|<\epsilon$. Here we have $f(x) = \dfrac{3x^{1/3}}{x+4}, a = 8$, and $L = \dfrac{1}{2}$. Substituting these into the inequalities in the limit definition and simplifying yields
$8-\delta < x < 8 + \delta$ and $|\dfrac{6x^{1/3}-(x+4)}{2(x+4)}|<\epsilon$.
We multiply both the numerator and denominator of the expression $|\dfrac{6x^{1/3}-(x+4)}{2(x+4)}|$ by $36x^{2/3}+6x^{1/3}+(x+4)^2$ to get $\dfrac{216x-(x+4)^3}{2(x+4)(36x^{2/3}+6x^{1/3}+(x+4)^2}$. Let $\delta \leq 1$. Then $7 < x < 9$ so $2(x+4)(36x^{2/3}+6x^{1/3}+(x+4)^2 > 2(7+4)+36(3)+6(1)+(7+4)^2 = 257$. Thus, we have that $\dfrac{216x-(x+4)^3}{2(x+4)(36x^{2/3}+6x^{1/3}+(x+4)^2}$ < $\dfrac{216x-(x+4)^3}{257}$ < $\dfrac{216(8+\delta)-(8-\delta+4)^3}{257}$. Simplifying, $\dfrac{216(8+\delta)-(8-\delta+4)^3}{257} = \dfrac{216\delta+432\delta-3\delta^2+\delta^3}{257} = \dfrac{\delta(\delta^2-3\delta+648)}{257} = \dfrac{\delta(\delta^2-3\delta+648)}{257}=\dfrac{\delta(\delta-\dfrac{3}{2})^2+645.75\delta}{257}<\dfrac{648\delta}{257}$ $\dfrac{}{}$ (as $1 \geq\delta >0$), and so let $\delta = \min\{1,\dfrac{257}{648}\epsilon\}$. Then we have $\dfrac{648\delta}{257} \leq \dfrac{648*\dfrac{257\epsilon}{648}}{257}=\epsilon$. Okay, so I'm not sure whether I made any mistakes and I feel that this method is very long-winded. Is there a shorter way to show this?
*For some reason, the code doesn't show up right when others view this question, but it does for me. I have no idea why.
Best Answer
First note that $|x^{1/3}-2|=\frac {|x-8|}{|x^{2/3}+2x^{1/3}+4|}<\frac{|x-8|}{11}$, for $x>7.$
Given $\epsilon >0$, let $|x-8|< \min (\epsilon,1)$. Then, as required,
$$|\dfrac{3x^{1/3}}{x+4}- \dfrac{1}{2}|=\frac {|6x^{1/3}-12+8-x|}{2(x+4)}\le \frac{6}{22}|x^{1/3}-2|+\frac{1}{22}|x-8|<\epsilon .$$