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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.

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    $\begingroup$ You are SUBTRACTING $x^2$. This term is the highest order so it will dwarf everything else in your limit. $\endgroup$
    – graydad
    Sep 15, 2014 at 17:31
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    $\begingroup$ You were so close! Check the sign of your $x^2$ term :) $\endgroup$
    – fixedp
    Sep 15, 2014 at 17:32
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    $\begingroup$ $\lim_{x\rightarrow +\infty} x^2 \left( \frac{\sin(x)}{x} + \frac{\cos(x)}{x^2} - 1 \right)$ is equivalent to $\lim_{x\rightarrow +\infty} -x^2 \left( -\frac{\sin(x)}{x} - \frac{\cos(x)}{x^2} + 1 \right)$ $\endgroup$
    – Amateur
    Sep 15, 2014 at 17:33

3 Answers 3

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

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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 $\infty$

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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.

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