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'.