Question

Find $\underset{x\to -\infty }{lim}\frac{x^4\sin \left(1/x\right)+x^2}{1+{\left|x\right|}^3}$

Answer 1

$\underset{x\to -\infty }{lim}\frac{x^4\sin \left(1/x\right)+x^2}{1+\left|x^3\right|}$
Substitute $x=-\frac{1}{y}\therefore y\to 0$ and $y$ is +ive when $x\to -\infty$
$=\underset{y\to 0}{lim}\frac{\left(1/y^4\right)\sin (-y)+\left(1/y^2\right)}{1+{\left|(-1/y)\right|}^3}$
$=\underset{y\to 0}{lim}\frac{(y^2-\sin y)/y^4}{|1+\left(1/y^3\right)|}$
$\because {|-\frac{1}{y}|}^3=\frac{{|-1|}^3}{{\left|y\right|}^{{}^3}}=\frac{1}{y^3}(y>0)$
$=\underset{y\to 0}{lim}\frac{y^2-\sin y}{y^4+y}\left(Form\frac{0}{0}\right)$
using L' hospital method
$=\underset{y\to 0}{lim}\frac{2y-\cos y}{4y^3+1}=\frac{0-1}{0+1}=-1.$