<s>[INST] Computing $\lim_{x\to+\infty}\frac{(x^{3/2}-2x+1)^{2/3}-x}{\sqrt{x+1}} $ $$
\lim_{x\to+\infty}\frac{(x^{3/2}-2x+1)^{2/3}-x}{\sqrt{x+1}}
$$
Any ideas, guys? Can't use De L'Hospital yet, unluckily
Thanks
 [/INST] Let's get rid of the $2/3$ exponent. Since
$$a^3-b^3 = (a-b)(a^2 + ab + b^2)$$
Let $a=(x^{3/2}-2x+1)^{2/3}$, $b=x$, then
$$\frac{(x^{3/2}-2x+1)^{2/3}-x}{\sqrt{x+1}} = \frac{a-b}{\sqrt{x+1}}\frac{a^2+ab+b^2}{a^2 +ab+b^2} = \frac{a^3-b^3}{\sqrt{x+1}(a^2 + ab+ b^2)}$$


*

*$a^3-b^3$ simplifies to:
$$-4x^{5/2} + 4x^2 + 2x^{3/2} - 4x + 1$$
which is like -4$x^{5/2}$ when $x$ is large.

*$a^2 + ab+ b^2$ is:
$$(x^{3/2}-2x+1)^{4/3} + (x^{3/2}-2x+1)^{2/3} x + x^2 \sim 3x^2 $$
and thus this part is like $3x^2$ when $x$ is large.


Then we can write

$$\lim_{x\to+\infty}\frac{(x^{3/2}-2x+1)^{2/3}-x}{\sqrt{x+1}} = \lim_{x\to+\infty} \frac{-4x^{5/2}}{3x^2\sqrt{x+1}} = -\frac{4}{3}$$

</s>