Limit approaching to negative infinity. Q. Find $\lim _{x\to -\infty }\left(\frac{x^4\sin\frac{1}{x}+x^2}{1+|x|^3}\right)$
By inserting $x=-\frac{1}{y}$ and as $_{x\to \:-\infty \:}$ then $_{y\to \:0\:\:}$. By applying this my text arrive at an answer of -1.
But instead if we insert $x=\frac{1}{y}$, still when $_{x\to \:-\infty \:}$ we have $_{y\to \:0\:\:}$.
So we have the limit as 
$\lim _{x\to -\infty }\left(\frac{\sin y+y^2}{y^4\left(1+\frac{1}{\left|y\right|^3}\right)}\right)=\lim \:_{y\to \:0\:}\left(\frac{\sin y+y^2}{y^4+\left|y\right|}\right)=\lim \:_{y\to \:0\:}\left(\frac{\cos y+2y}{4y^3+1}\right)=\frac{\left(1+0\right)}{\left(0+1\right)}=1$
Which is negative of the actual answer. Am I wrong anywhere in the above method, I doubt my assumptions about the absolute 'x'. 
 A: If $x= -\frac 1 y$ then $y=-\frac 1 x$ so $|y|=y$ because $x \to -\infty \Rightarrow x<0 \Rightarrow y>0$.
If however, $x= \frac 1 y$, with the same reasoning as before we can conclude that $|y|=-y$.
Note that for each of the previous substitutions, $y \to 0$ from different sides. In the first case, y is positive and decreasing in value as $x  \to -\infty$, which can be noted as $y \to 0^+$. In the second case, the opposite is true, ie $y \to 0^-$. That is why $|y|$ behaves differently in each case.
Best solution, though, is to substitute $|x|=-x$ as $x\to -\infty$ before substituting $x=\pm \frac 1 y$. This is simpler, so less error prone.
A: Hint: If $y=x^{-1}$, then $$y\to 0^-$$
(that is, approaches $0$ while negative), and 
$$|y| = -y \quad {\rm for} \quad y<0.$$
Then use L'Hospital for left-hand limits.
$$\lim_{y\to 0^-}\left(\frac{\sin y+y^2}{y^4\left(1+\left|y\right|^{-3}\right)}\right)=\lim_{y\to 0^-}\left(\frac{\sin y+y^2}{y^4 - y}\right)=\lim_{y\to 0^-}\left(\frac{\cos y+2y}{4y^3 - 1}\right) =-1 $$
