Proving the limit $\lim_{x \to -2} \frac{\sqrt[3]{x}+\sqrt[3]{2}}{x+2}=\frac{\sqrt[3]{2}}{6}$ using the epsilon-delta definition.

Question

Prove $\lim_{x \to -2} \frac{\sqrt[3]{x}+\sqrt[3]{2}}{x+2}=\frac{\sqrt[3]{2}}{6}$.

Solution Attempt:

$\left| f(x) - L \right| < \epsilon$

$\left|f(x)-L \right|=\left| \frac{\sqrt[3]{x}+\sqrt[3]{2}}{x+2} - \frac{\sqrt[3]{2}}{6} \right|$

$=\left| \frac{6\left( \sqrt[3]{x}+\sqrt[3]{2}\right)-\sqrt[3]{2} (x+2)}{6(x+2)} \right|$

$=\left| \frac{6\sqrt[3]{x}+4\sqrt[3]{2}-\sqrt[3]{2}x}{6(x+2)}\right|$

$=\frac{1}{6}\left| \frac{6\sqrt[3]{x}+4\sqrt[3]{2}-\sqrt[3]{2}x}{x+2}\right|$

$\left| \frac{6\sqrt[3]{x}+4\sqrt[3]{2}-\sqrt[3]{2}x}{x+2}\right|<6\epsilon$

I am now stuck at this part of the solution. Can someone assist me?

Answer 1

For any $\,x\in(-4,16)\,\land\,x\neq-2\,$ it results that

$\dfrac16\,\left|\dfrac{6\sqrt[3]x\!+\!4\sqrt[3]2\!-\!\sqrt[3]2\,x}{x+2}\right|=\dfrac16\,\left|\dfrac{\sqrt[3]2\left(2\sqrt[3]2\!-\!\sqrt[3]x\right)\!\left(\sqrt[3]x\!+\!\sqrt[3]2\right)^2}{x+2}\right|\!=$

$\!\!\!\!\!\!\!=\!\dfrac16\dfrac{\sqrt[3]2\!\left(2\sqrt[3]2\!-\!\sqrt[3]x\right)\!\!\left(\sqrt[3]x\!+\!\sqrt[3]2\right)^2}{\left|x+2\right|}\!<\!\dfrac16\dfrac{\sqrt[3]2\!\left(2\sqrt[3]2\!+\!\sqrt[3]4\right)\!\!\left(\sqrt[3]x\!+\!\sqrt[3]2\right)^2}{\left|x+2\right|}\!=$

$=\dfrac16\,\dfrac{\sqrt[3]2\!\left(2\sqrt[3]2+\sqrt[3]4\right)\!\left(\sqrt[3]x+\sqrt[3]2\right)^2\left(\sqrt[3]{x^2}-\sqrt[3]{2x}+\sqrt[3]4\right)^2}{\left|x+2\right|\!\left(\sqrt[3]{x^2}-\sqrt[3]{2x}+\sqrt[3]4\right)^2}=$

$\!\!\!\!\!\!=\dfrac16\dfrac{\sqrt[3]2\!\left(2\sqrt[3]2\!+\!\sqrt[3]4\right)\!\left(x\!+\!2\right)^2}{\big|x\!+\!2\big|\,\left[\left(\!\sqrt[3]x\!-\!\frac{\sqrt[3]2}2\right)^2\!\!+\!\frac{3\sqrt[3]4}4\right]^2}=\dfrac16\dfrac{\sqrt[3]2\!\left(2\sqrt[3]2\!+\!\sqrt[3]4\right)\big|x\!+\!2\big|}{\left[\left(\!\sqrt[3]x\!-\!\frac{\sqrt[3]2}2\right)^2\!\!+\!\frac{3\sqrt[3]4}4\right]^2}\leqslant$

$\leqslant\dfrac16\,\dfrac{\sqrt[3]2\!\left(2\sqrt[3]2+\sqrt[3]4\right)\big|x+2\big|}{\left(\frac{3\sqrt[3]4}4\right)^2}=\dfrac16\,\dfrac{\sqrt[3]2\!\left(2\sqrt[3]2+\sqrt[3]4\right)\big|x+2\big|}{\frac{9\sqrt[3]2}8}<$

$<\dfrac16\,\dfrac{\sqrt[3]2\big(4+2\big)\,\big|x+2\big|}{\frac{9\sqrt[3]2}8}=\dfrac89\,\big|x+2\big|<\big|x+2\big|\,.$

We have proved that

$\dfrac16\,\left|\dfrac{6\sqrt[3]x+4\sqrt[3]2-\sqrt[3]2\,x}{x+2}\right|<\big|x+2\big|\;\,,\;\;\,\forall x\!\in\!(-4,16)\setminus\{-2\}.$

Consequently,

for any $\,\varepsilon>0\,$ there exists $\,\delta=\min\{2,\varepsilon\}>0\,$ such that

$\forall x\!\in\!\big(\!\!-\!2\!-\!\delta,-2\!+\!\delta\big)\!\setminus\!\{-2\}\subsetneq(-4,16)\!\setminus\!\{-2\}\,$ it results that

$\big|f(x)-L\big|=\dfrac16\,\left|\dfrac{6\sqrt[3]x+4\sqrt[3]2-\sqrt[3]2\,x}{x+2}\right|<\big|x+2\big|<\delta\leqslant\varepsilon\,.$

Answer 2

Since $x = y^3$ is bijection on $\mathbb{R}$, applying a change of variable makes it much easier $$ \lim_{x\to -2}\frac{\sqrt[3]{x} + \sqrt[3]{2}}{x + 2}\\ \lim_{y\to -\sqrt[3]{2}}\frac{1}{y^2 - \sqrt[3] 2y + 2^{2\over3}} $$

Apparently $\lim_{y\to -\sqrt[3]{2}}{y^2 - \sqrt[3] 2y + 2^{2\over3}}\ne 0$, you could apply $\epsilon-\delta$ definition on this and trace back to original limit.