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

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.
