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
 A: 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

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

A: HINT:
Set $\sqrt x=y$
$$\lim_{x\to+\infty}\frac{(x^{3/2}-2x+1)^{2/3}-x}{\sqrt{x+1}}$$
$$=\lim_{y\to+\infty}\frac{(y^3-2y^2+1)^{2/3}-y^2}{\sqrt{y^2+1}}$$
Use  $a^3-b^3=(a-b)(a^2+ab+b^2)$ for
$(y^3-2y^2+1)^{2/3}-y^2=\{(y^3-2y^2+1)^2\}^{\frac13}-(y^6)^{\frac13}$
$$=\frac{(y^3-2y^2+1)^2-y^6}{\{(y^3-2y^2+1)^2\}^{\frac23}+\{(y^3-2y^2+1)^2\}^{\frac13}y^2+(y^2)^2}$$
$$=\frac{(-2y^2+1)^2+2y^3(-2y^2+1)}{\{(y^3-2y^2+1)^2\}^{\frac23}+\{(y^3-2y^2+1)^2\}^{\frac13}y^2+(y^2)^2}$$
Hope you can take it home from here
A: In problems such as this one, the trick is to "clean up" the different terms. Here you should start with the first term in the numerator. Identify the leading term for   $x\rightarrow \infty$. This is $x^{3/2}$. Take it out of the term between brackets. You will see that the $\frac 3 2$ and $\frac 2 3$ powers cancel. The remaining term between brackets behaves nicely in the limit $x\rightarrow \infty$. So you can expand it in a Taylor series. This will lead you to the following result. The numerator becomes $x[1-\frac{(4/3)}{\sqrt x}+.....] -x$. After cancellation of the terms with x we see that the leading term in the numerator goes as $-(4/3)\sqrt x$. Hence the quotient of numerator and denominator has the limit: $-\frac 4 3$.
@Ludovico L : I see that you have received two other replies to your question. In both cases the highly learned respondents come up with an overly elaborate method. They both have spotted that the first term in the numerator equals zero for $x=1$. Yihoo! How clever! To exploit this little fact, they come up with some ingenious substitution. Well, one can obviously do that; and it might even be useful for some other (analytical) purposes! But to find the limit of your function, which is actually a very simple procedure, it is complete and utter overkill. The appropriate procedure is to identify and isolate the leading term, as I explained in the first part of my reply. This generally works, and has very little to do with accidental properties of the function at hand. To illustrate this, lets see what happens if we further mess up the first term in the numerator of your function $f(x)$:
$f(x) = \frac N D$
$N = (x^{3/2} - 2x + 666 - \frac 3 x + e^{-x} + 7\sin(x))^{2/3} - x $
$D = \sqrt{x + 1}$
Going through the recommended procedure, we can swiftly establish that this function has exactly the same limit for $x\rightarrow +\infty$ as your function: $-\frac 4 3 $. Such is the power of analysis, and in particular of the universal method of Taylor series expansion. In my opinion, that is what this exercise was all about.     
