Calculating $\lim_{x\rightarrow +\infty} x \sin(x) + \cos(x) - x^2 $

Question

I want to calculate the following limit:

$\displaystyle\lim_{x\rightarrow +\infty} x \sin(x) + \cos(x) - x^2 $

I tried the following:

$\displaystyle\lim_{x\rightarrow +\infty} x \sin(x) + \cos(x) - x^2 = $

$\displaystyle\lim_{x\rightarrow +\infty} x^2 \left( \frac{\sin(x)}{x} + \frac{\cos(x)}{x^2} - 1 \right) = +\infty$

because

$\displaystyle\lim_{x\rightarrow +\infty} x^2 = +\infty$,

$\displaystyle\lim_{x\rightarrow +\infty} \frac{\sin(x)}{x} = 0$, and

$\displaystyle\lim_{x\rightarrow +\infty} \frac{\cos(x)}{x^2} = 0$,

but I know from the plot of the function that this limit goes to $- \infty$, so I'm clearly doing something wrong. Sorry in advance for the simple question and perhaps for some silly mistake.

Answer 1

Using your work: $$\lim_{x\rightarrow +\infty} x^2 \left( \underbrace{\frac{\sin(x)}{x}}_{\to 0} + \underbrace{\frac{\cos(x)}{x^2}}_{\to 0} \color{blue}{\bf - 1 }\right) = \lim_{x\rightarrow +\infty} x^2(-1) = -\infty$$

Answer 2

Denote $f(x)$ the given expression then since $\sin$ and $\cos$ functions are bounded then we have easily

$$\lim_{x\to\infty}\frac{f(x)}{x^2}=-1\implies f(x)\sim_\infty-x^2$$ which means that $f(x)$ is asymptotically equivalent to $-x^2$ at $\infty$

Answer 3

Hint: $-\frac{1}{2}x^2\geq x+1-x^2\geq (x\sin x+\cos x-x^2)$ for $x\,$ large enough. Now let $x$ go to infinity.