Evaluation of $2$ limit problems Evaluation of $(a)\;\;\lim_{x\rightarrow \infty}\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
and $(b)\;\; \displaystyle \lim_{x\rightarrow 0^{-}}\frac{\lfloor x \rfloor +\lfloor x^2 \rfloor+...............+\lfloor x^{2n+1} \rfloor +n+1}{1+\lfloor x^2 \rfloor +|x|+2x}.$
$\bf{My\; Try::(a):}$ Given $\lim_{x\rightarrow \infty}\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
Now Multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\left\{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$.
So $\displaystyle \lim_{x\rightarrow \infty}\frac{\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}}{\left\{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}\right\}}\times \left\{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
So we Get $\displaystyle \lim_{x\rightarrow \infty} \frac{(a-2)x^3+(3-c)x^2+(b-3)x+(2-d)}{x^2\cdot \left(\sqrt{1+\frac{a}{x}+\frac{3}{x^2}+\frac{b}{x^3}+\frac{2}{x^4}}+\sqrt{1+\frac{2}{x}-\frac{c}{x^2}+\frac{3}{x^3}-\frac{d}{x^4}}\right)}$
Now If limit is finite , Then $(a-2)=0\Rightarrow a=2$
can we solve the first question is any other way, If yes the plz explain me,
and how can i solve $(2)$ one.
Thanks
 A: Assuming $\;\lfloor x\rfloor\;$ is the floor function, we get that for $\;x\;$ very close to zero from the left, 
$$\;\forall\,k\in\Bbb N\;,\;\;\lfloor x^k\rfloor=\begin{cases}-1&,\;\;k\;\;\text{is odd}\\{}\\\;\;\,0&,\;\;k\;\;\text{is even}\end{cases}\;$$ , and then:
$$\frac{\lfloor x \rfloor +\lfloor x^2 \rfloor+\ldots+\lfloor x^{2n+1} \rfloor +n+1}{1+\lfloor x^2 \rfloor +|x|+2x}=\frac{\overbrace{-1+0-1+\ldots+0-1}^{\lfloor x\rfloor+\ldots+\lfloor x^{2n+1}\rfloor}+n+1}{1+0-1+2x}=$$
$$=\frac{-1(n+1)+n+1}{2x}=0\xrightarrow[x\to 0^-]{}0$$
A: Not a full answer, just expanding my comment about $(a)$.
We can rewrite the function (keeping in mind that $x \to \infty$) like this:
\begin{align}
f(x)&:=\sqrt{x^4 + ax^3 + 3x^2 + bx + 2} - \sqrt{x^4 + 2x^3 -cx^2 + 3x - d} =\\
&=x^2\left(1 + \frac a{2x} + \frac 3{2x^2} + \frac b{2x^3} + \frac 1{x^4} - 1 - \frac 1x + \frac c{x^2} - \frac 3{2x^3} + \frac d{2x^4} + \mathcal{o}\left(\frac 1{x^4}\right)\right)=\\
&=\left(\frac a2 - 1\right)x + \frac 32 + c + \mathcal{o}(1)
\end{align}
Now it's easy to see that
$$\lim_{x \to \infty} f(x) = \begin{cases}\frac 32 + c&\qquad a = 2\\
\infty&\qquad a \neq 2
\end{cases}$$
I've made use of the expansion:
$$\sqrt{1 + t} = 1 + t/2 + \mathcal{o}(t)\qquad\text{for}\ t \to 0$$
