Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$

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 $$\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}} = \frac 12 \cdot\lim_{t \to 0+} \frac{f(t)-f(-t)}{t} \cdot \lim_{t \to 0+} \frac{2t}{f(2t)-f(-2t)} = \frac 12 \cdot \frac 23 \cdot \frac 32 = \frac 12$$

From the binomial theorem, we know $(1+y)^n\approx1+ny$ for $y\approx0$. Divide both numerator and denominator by $x^{2/3}$. The limit is$$\lim_{x \to \infty}{\frac{\left(1+\frac1x\right)^{\frac{2}{3}}-\left(1-\frac1x\right)^{\frac{2}{3}}}{\left(1+\frac2x\right)^{\frac{2}{3}}-\left(1-\frac2x\right)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{1+\frac{2/3}x-1+\frac{2/3}x}{1+\frac{4/3}x-1+\frac{4/3}x}=1/2.$$

Here is how I would do it without using L'hopital, or series expansion :

Using the identity $ a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right) $, we have : $$ \left(x+1\right)^{2}-\left(x-1\right)^{2}=\left(\left(x+1\right)^{\frac{2}{3}}-\left(x-1\right)^{\frac{2}{3}}\right)\left(\left(x+1\right)^{\frac{4}{3}}+\left(x^{2}-1\right)^{\frac{2}{3}}+\left(x-1\right)^{\frac{4}{3}}\right) $$

And : $$ \left(x+2\right)^{2}-\left(x-2\right)^{2}=\left(\left(x+2\right)^{\frac{2}{3}}-\left(x-2\right)^{\frac{2}{3}}\right)\left(\left(x+2\right)^{\frac{4}{3}}+\left(x^{2}-4\right)^{\frac{2}{3}}+\left(x-2\right)^{\frac{4}{3}}\right) $$

Thus : \begin{aligned}\lim_{x\to +\infty}{\frac{\left(x+1\right)^{\frac{2}{3}}-\left(x-1\right)^{\frac{2}{3}}}{\left(x+2\right)^{\frac{2}{3}}-\left(x-2\right)^{\frac{2}{3}}}}&=\lim_{x\to +\infty}{\frac{\left(\left(x+1\right)^{2}-\left(x-1\right)^{2}\right)\left(\left(x+2\right)^{\frac{4}{3}}+\left(x^{2}-4\right)^{\frac{2}{3}}+\left(x-2\right)^{\frac{4}{3}}\right)}{\left(\left(x+2\right)^{2}-\left(x-2\right)^{2}\right)\left(\left(x+1\right)^{\frac{4}{3}}+\left(x^{2}-1\right)^{\frac{2}{3}}+\left(x-1\right)^{\frac{4}{3}}\right)}}\\ &=\lim_{x\to +\infty}{\frac{\left(1+\frac{2}{x}\right)^{\frac{4}{3}}+\left(1-\frac{4}{x^{2}}\right)^{\frac{2}{3}}+\left(1-\frac{2}{x}\right)^{\frac{4}{3}}}{2\left(\left(1+\frac{1}{x}\right)^{\frac{4}{3}}+\left(1-\frac{1}{x^{2}}\right)^{\frac{2}{3}}+\left(1-\frac{1}{x}\right)^{\frac{4}{3}}\right)}}\\ &=\frac{1}{2}\end{aligned}