$\varepsilon$ - $\delta$ proof for $\lim_{x \to 27}2x^{2/3}=18$

Question

Construct a careful $\varepsilon$ - $\delta$ argument to show $$\lim_{x \to 27}2x^{2/3}=18$$

From the definition of a limit $$\forall \varepsilon > 0, \space \exists \delta >0 \space : \space 0<\left| x-a \right| < \delta \implies \left| f(x) - L \right| < \varepsilon$$ I've rearranged as follows $$\left| 2x^{2/3}-18 \right|<\varepsilon$$ $$\left| 2 \right | \left| x^{2/3}-9 \right|<\varepsilon$$ $$\left| x^{2/3}-9 \right|< \frac{\varepsilon}{2}$$ $$\left| x^{1/3}-3 \right| \left| x^{1/3}+3 \right| < \frac{\varepsilon}{2}$$ $$\left| x^{1/3}-3 \right| < \frac{\varepsilon}{2(x^{1/3}+3)}$$ $$\left| x-9x^{2/3}+27x^{1/3}-27 \right| < \left( \frac{\varepsilon}{2(x^{1/3}+3)} \right)^3$$ $$\left| (x-27)-9x^{2/3}+27x^{1/3} \right| < \left( \frac{\varepsilon}{2(x^{1/3}+3)} \right)^3$$

so here I have $\left| x-a\right|$ on the left but I'm not sure how to get to constructing $\delta$ from this expression.

Answer 1

$$\left| x^{1/3}-3 \right| < \frac{\varepsilon}{2|x^{1/3}+3|}$$ $$\left|\frac {x-27}{(x^{1/3}-3)^2+9x^{1/3}} \right| < \frac{\varepsilon}{2|x^{1/3}+3|}$$ $$\left|{x-27} \right| < \frac{\varepsilon|{(x^{1/3}-3)^2+9x^{1/3}}|}{2|x^{1/3}+3|}$$

Because $a^3-b^3=(a-b)^3+3ab(a-b)$ where $a=x^{1/3}$ and $b=3$ in this case. $$I.e. \space \space x^{1/3}-3=\frac {x-27}{(x^{1/3}-3)^2+9x}$$

after which you can simplify by introducing $\left|x-27\right|<1$

Im not sure if i'm right but this is what i would do.Hope it helps

Edit: was that clear enough? P.S. I'm still new to the mathjax coding stuff. hopefully there're no mistakes here

Answer 2

Since $x$ is near $27$ there's no harm in mandating $\delta<1$ so that $\lvert x-27\rvert<\delta$ and $\lvert x+27\rvert <55$. Now, \begin{equation} \lvert x^{2/3} - 9\rvert = \lvert x^{2/3} - 27^{2/3}\rvert = \frac{\lvert x^2-27^2\rvert}{\lvert x^{4/3} + 27^{2/3}x^{2/3}+27^{4/3}\rvert} = \frac{\lvert x+27\rvert\lvert x-27\rvert}{\lvert x^{4/3} + 27^{2/3}x^{2/3}+27^{4/3}\rvert};\end{equation} the function in the denominator is bounded above on small intervals containing $27$, say by $M$, so $$\lvert x^{2/3}-9\rvert <\frac{55\delta}M.$$ For $\varepsilon>0$ given choose $\delta = \min\{1,\varepsilon M/55\}$.

I've left out some algebra and other details (including the original factor of 2), but this should get you started.

Answer 3

At the fourth step, use $(a^3-b^3) = (a-b)(a^2+ab+b^2)$

$$\begin{align} |2x^{2/3}-18| &= 2\left| x^{1/3}-3 \right| \left| x^{1/3}+3 \right|\\ & = 2\frac {\left| x-27 \right|\left|x^{1/3}+3\right|}{\left|x^{2/3}+3x^{1/3}+9\right|} \\ & <2\left| x-27 \right|\left|x^{1/3}+3\right| \end{align} $$

Now restrict $\delta \leq 1$. We are doing this so that we can make $\left|x^{1/3}+3\right|$ bounded.
Since $$\begin{align} |x| - |27| \leq | x-27| &<1 \\ \implies |x| &< 28 \\ \end{align}$$

$$\begin{align} 2\left|x^{1/3}+3\right| &\leq 2(\left|x\right|^{1/3}+3) \\ & < 2(28^{1/3} + 3)\\ & = M \end{align}$$

So, finally you get $|2x^{2/3}-18| < M|x-27| < \varepsilon \iff |x-27| < \delta := \min\{\frac {\varepsilon}{M}, 1\}$

Answer 4

Hint:

$\left\lvert x-27\right\rvert<\epsilon\\ \implies 27-\epsilon<x<27+\epsilon\\ \implies 3-\epsilon^{1/3}<\sqrt[3]{27-\epsilon}<\sqrt[3]{x}<\sqrt[3]{27+\epsilon}<3+\epsilon^{1/3} \ (\forall \epsilon<3)$

Answer 5

You need to establish $$-\delta<x-27<\delta\implies -\epsilon<2x^{2/3}-18<\epsilon \\\equiv\left(\frac{18-\epsilon}2\right)^{3/2}<x<\left(\frac{18+\epsilon}2\right)^{3/2} \\\equiv\left(\frac{18-\epsilon}2\right)^{3/2}-27<x-27<\left(\frac{18+\epsilon}2\right)^{3/2}-27.$$

An obvious choice for $\delta$ is the smallest of $$27-\left(\frac{18-\epsilon}2\right)^{3/2}\text{ and }\left(\frac{18+\epsilon}2\right)^{3/2}-27.$$ By monotonicity of the power, this expression is strictly positive and defines a suitable $\delta$ for any $\epsilon$.

[For completeness: when $\epsilon>18$ the left bound is undefined, just use the right one.]

More generally, for an increasing function $f$, $$-\delta<x-a<\delta\implies-\epsilon<f(x)-L<\epsilon \\\equiv f^{-1}(L-\epsilon)-a<x-a<f^{-1}(L+\epsilon)-a,$$ and $$\delta=\text{min}\left(a-f^{-1}(L-\epsilon),f^{-1}(L+\epsilon)-a\right)$$ will do if you can show that it is defined and positive.