# Does $x\cos(x)$ have oblique asymptotes?

Looking at the graph of $x\cos(x)$ or $x\sin(x)$ etc., it looks like the magnitude of the waves are following a line. Are they oblique asymptotes or something else?

I am familiar with finding the oblique asymptotes of a rational function like $\frac{P(x)}{Q(x)}$ by dividing $Q(x)$ into $P(x)$; however, it doesn't seem like I can do that with $x \cos(x)$ et al. what is going on here? and how would you find the equation of the function that is 'controlling' the original function's behavior?

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Btw: The magnitudes are following a line, namely the line $y=x$ (for sin). Just the term oblique asymptote doesn't fit here. – Simon Markett Jun 12 '12 at 10:17
How do you know the magnitudes follow the line $y = x$? – stariz77 Jun 12 '12 at 10:24
Well, if you look at $x\sin(x)$, then $\sin(x)$ oscillates between $\pm 1$ and $x$ is, well, $x$. So the magnitues are attended whenever $x\sin(x)=\pm x$. Hence they are on the lines $x=y$ and $y=-x$. The same holds for $\cos$. – Simon Markett Jun 12 '12 at 10:28
Even less than $\cos x$ has a horizontal one. – mrf Jun 12 '12 at 10:43

For a line to count as an asymptote, the distance between that line and the graph of the function has to tend to zero. This doesn't hold in your examples, so the lines are not asymptotes.

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So, what are they? Can you derive them from the original equation? – stariz77 Jun 12 '12 at 10:16
I don't know if they have a name, but since $\cos x$ oscillates between $-1$ and $+1$, clearly the curve $y = x \cos x$ will oscillate between the lines $y=-x$ and $y=+x$. – Hans Lundmark Jun 12 '12 at 10:18

Hint: If you are to have an oblique asymptote, then you need a linear function $y = ax+b$ such that $\lim_{x\rightarrow\infty} \frac{x\sin x}{ax+b}$ exists. Can such a linear function exist?

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Do you mean $x\to\infty$? – Simon Markett Jun 12 '12 at 10:13
$$\lim_{x\rightarrow \infty } \frac{x\sin x}{ax+b} = 0 ???$$ ?? – Santosh Linkha Jun 12 '12 at 10:17
Already corrected the limit to infinity and not to zero – DonAntonio Jun 12 '12 at 10:36
@Experiment No, that limit is not zero...! To see this easily, take $\,b=0\,$...does the limit exist? Now convince yourself that adding $\,b\,$ in the denominator changes nothing really important. – DonAntonio Jun 12 '12 at 10:38