Evaluate $\lim_{x\to -\infty} \frac{x^4\sin( \frac 1x )+ x^2}{1+|x|^3}$

Question

Let $y=-x$ $$\lim_{y\to \infty} \frac{y^4 \sin (-\frac 1y) + y^2}{1+|y|^3}$$ $$=\lim_{y\to \infty} \frac{y^3 ((-y \sin \frac 1y) +\frac 1y)}{y^3(1+\frac{1}{y^3})}$$

At $y\to \infty$, $y\sin \frac 1y=0$ (What I think is happening)

Therefore the entire limit is $0$

I pointed out where I think I went wrong, but I am not able to correct it. What should be the right procedure?

Answer 1

Let $$L=\lim_{x\to -\infty} \frac{x^4\sin( \frac 1x )+ x^2}{1+|x|^3}$$ $$\implies L=\lim_{x \to -\infty} \frac{x^4/x+x^2}{1-x^3}=\lim_{x\to -\infty}\frac{ 1+1/x}{1/x^3-1}=-1.$$

Answer 2

Let $-1/x=h\implies h\to0^+$

$$\lim_{h\to0^+}\dfrac{(\sin(-h) +h^2)h^3}{h^4(h^3+1)}=\lim_{h\to0^+}\dfrac1{h^3+1}\cdot\lim_{h\to0^+}\dfrac{-\sin h+h^2}h=?$$