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...
 A: 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 $$\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}} = \frac 12 \cdot\lim_{t \to 0+} \frac{f(t)-f(-t)}{t} \cdot \lim_{t \to 0+} \frac{2t}{f(2t)-f(-2t)} = \frac 12 \cdot \frac 23 \cdot \frac 32 = \frac 12$$
A: 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$$\lim_{x \to \infty}{\frac{\left(1+\frac1x\right)^{\frac{2}{3}}-\left(1-\frac1x\right)^{\frac{2}{3}}}{\left(1+\frac2x\right)^{\frac{2}{3}}-\left(1-\frac2x\right)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{1+\frac{2/3}x-1+\frac{2/3}x}{1+\frac{4/3}x-1+\frac{4/3}x}=1/2.$$
A: 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}
A: With difference of squares and difference of cubes, you get that the limit simplifies to
$$\lim_{x\to\infty} \frac{(x+1)-(x-1)}{(x+2)-(x-2)}\cdot \frac{f(x)}{g(x)} = \frac{1}{2}$$
because $f$ and $g$ both grow at the rate $\sim 6 x$ for large $x$.
A: Hint:
$$\lim_{x\to\infty}\dfrac{(x+1)^{2/3}-(x-1)^{2/3}}{(x+2)^{2/3}-(x-2)^{2/3}}=\dfrac12\cdot\lim_{t\to0}\dfrac{\dfrac{(1+t)^{2/3}-(1-t)^{2/3}}t}{\dfrac{(1+2t)^{2/3}-(1-2t)^{2/3}}{2t}}=\dfrac12\cdot\dfrac{\lim_{t\to0}f(t)}{\lim_{t\to0}f(2t)}$$
Now for $\lim_{t\to0}f(t)=\lim_{t\to0}\dfrac{(1+t)^{2/3}-(1-t)^{2/3}}t,$  set $(1+t)^{1/3}=p,(1-t)^{1/3}=q$
$\implies p^3-q^3=2t$ and $\implies p\to q, q\to1$
$$\lim_{t\to0}f(t)=\lim_{p\to q, q\to1}\dfrac{2(p-q)}{p^3-q^3}=\lim_{q\to1}\dfrac2{3q^2}=?$$
