Find limit of function $x^2 \sin{\frac{1}{x}}-x-xe^x$ as $x \rightarrow -\infty$

Question

Find limit of function $$x^2 \sin\left(\frac{1}{x}\right)-x-xe^x$$ as $x \rightarrow -\infty$.

The result is zero.

Without using Taylor's expansion, since it comes from middle exam of the first semester.

Answer 1

The last term tends to $0$ (you should be able to prove this, it is one of the standard limits associated with exponential function) so you need to deal with $x^{2}\sin(1/x)-x$ and then put $x=1/t$ to get $$\frac{\sin t - t} {t^{2}}$$ which can be easily shown to have limit $0$ as $t\to 0^{-}$. So the desired limit is $0$.