Indeterminate form?

$\begin{aligned} \mathfrak L & = \lim_{x \to \infty} \left( \sqrt[3]{8^x+3^x} - \sqrt{4^x-2^2} \right) \\ & = \lim_{x \to \infty} \left( 2^x \left[1+ \left(\frac{3}{8}\right)^x \right]^\frac{1}{3} - 2^x \left[1 - \left(\frac{2}{4}\right)^x \right]^\frac{1}{2} \right) \quad \quad \small \color{#3D99F6}{\text{By binomial expansion}} \\ & = \lim_{x \to \infty} \left(2^x \left[1+ \frac{1}{3}\left(\frac{3}{8}\right)^x - \frac{1}{9}\left(\frac{3}{8}\right)^{2x} + ... \right] - 2^x \left[1 - \frac{1}{2}\left(\frac{1}{2}\right)^x - \frac{1}{8}\left(\frac{1}{2}\right)^{2x} - ... \right] \right) \\ & = \lim_{x \to \infty} \left(\left[\color{#3D99F6}{2^x} + \color{#D61F06}{\frac{3^{x-1}}{2^{2x}} - \frac{3^{2x-2}}{2^{5x}} + ... } \right] - \left[\color{#3D99F6}{2^x} - \frac{1}{2} - \color{#D61F06}{\frac{1}{2^{x+3}} - ...} \right] \right) \quad \quad \small \color{#3D99F6}{2^x - 2^x = 0} \text{ and }\color{#D61F06}{\text{all red terms}\to 0 \text{ as }x \to \infty} \\ & = \frac{1}{2} = \boxed{0.5} \end{aligned}$

This problem is similar to Iikkyu San's problem. Waiting for Rishabh to post the answer.