Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{(\frac{x+1}{x-2})^{\frac{2}{3}}-(\frac{x-1}{x-2})^{\frac{2}{3}}}{(\frac{x+2}{x-2})^{\frac{2}{3}}-1}$$
L'Hopital
$$L=\lim_{x \to \infty}\frac{\frac{2}{3}(\frac{x+1}{x_2})^{-\frac{1}{3}}(\frac{x-2-(x+1)}{(x-2)^2})-\frac{2}{3}(\frac{x-1}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x-1)}{(x-2)^2})}{\frac{2}{3}(\frac{x+2}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x+2)}{(x-2)^2})}=\lim_{x\ \to \infty}{\frac{3(x+1)^{-\frac{1}{3}}-(-1)(x-1)^{-\frac{1}{3}}}{(x+2)^{-\frac{1}{3}}(-4)}}$$
$$=\lim_{x \to \infty}{\frac{3(1+\frac{1}{x})^{-\frac{1}{3}}-(1-\frac{1}{x})^{-\frac{1}{3}}}{4(1+\frac{2}{x})^{-\frac{1}{3}}}}=\frac{3-1}{(4)(1)}=\frac{1}{2}.$$
Is this correct and is there a more elegant way of doing it?
EDIT: strictly speaking L'Hopital is not applicable with $x \to \infty$ so just got lucky here...

Here is another way to do it. Dividing by $x^{2/3}$ yields $$ \frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}} =  \frac{(1+\frac 1x)^{\frac{2}{3}}-(1-\frac 1x)^{\frac{2}{3}}}{(1+\frac 2 x)^{\frac{2}{3}}-(1-\frac 2x)^{\frac{2}{3}}} 
$$
Substitute $t = 1/x$. Then $$\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}} = \lim_{t \to 0+}\frac{(1+ t)^{\frac{2}{3}}-(1-t)^{\frac{2}{3}}}{(1+ 2t)^{\frac{2}{3}}-(1- 2t)^{\frac{2}{3}}}  $$
Putting $f(t) = (1+t)^{2/3}$ yields \begin{align} \lim_{t \to 0+}\frac{(1+ t)^{\frac{2}{3}}-(1-t)^{\frac{2}{3}}}{(1+ 2t)^{\frac{2}{3}}-(1- 2t)^{\frac{2}{3}}} &= \lim_{t \to 0
+} \frac{f(t)-f(-t)}{f(2t)-f(-2t)} \\ &= \frac 12 \cdot\lim_{t \to 0+} \frac{f(t)-f(-t)}{t} \cdot \lim_{t \to 0+} \frac{2t}{f(2t)-f(-2t)}\end{align}
Note that $$ \lim_{t \to 0+} \frac{f(t)-f(-t)}{t}= \lim_{t \to 0+} \frac{f(t)-f(0)}{t} + \lim_{t \to 0+} \frac{f(-t)-f(0)}{-t} = 2f'(0)=\frac 23$$
Thus $$\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}} = \frac 12 \cdot\lim_{t \to 0+} \frac{f(t)-f(-t)}{t} \cdot \lim_{t \to 0+} \frac{2t}{f(2t)-f(-2t)} = \frac 12 \cdot \frac 23 \cdot \frac 32 = \frac 12$$
