The function is

$f(x) = \frac{\sin(x) + \cos (x) -1}{x + arctan(x)}$ When $x \neq 0$ and $\frac{1}{2}$ when $x = 0$

I know that to get vertical asymptote then

$\lim_{x \to a} f(x) = \pm \infty$

And when $y = g(x)$ is the other asymptote for f when ${x \to \infty}$ if

$\lim_{x \to \infty} (f (x) − g(x)) = 0$

The line is horisontal if

y = g(x) = L And angular if

$y = g(x) = ax + b, a \neq 0$.

However, I don't know how to make the function go to infinity, if I plug in zero, it's $\frac{0}{0}$. Is that considered infinity? How should I go with this?

  • $\begingroup$ Here it is helpful to know the ranges of the constituent functions. You know that for any input $x$, both sine and cosine are bounded by $y=\pm 1$. So as $x$ gets large, sine and cosine stay between $-1$ and $+1$. Also, $arctan(x)$ takes on values between $-\pi/2$ and $\pi/2$, so it is also bounded for any value of $x$, especially large values... Now, that leaves an $x$ in the denominator....... $\endgroup$ – Eleven-Eleven Oct 12 '18 at 12:52

Making the function "go to infinity" means we are interested in the behavior of the outputs when the inputs get extremely large. For a fixed numerator $k$ and positive rational number $r$, we have the following to help.


Essentially, by fixing the numerator and allowing the denominator to increase without bound, the value of the fraction then gets smaller to the point where it is essentially equal to $0$. You can think of this as indirect variation. As your denominator increases, the fraction value decreases. This also works in the "opposite" direction in that if your denominator decreases, the fraction values increase.

That is important because we have a power of $x$ in the denominator.

Now, your initial function has in its numerator the sine and cosine functions. Knowing the ranges of these are important. Both sine and cosine will never take on function values that are larger than $1$ or smaller than $-1$. Thus by adding them together their sum can never be greater than $2$ or less than $-2$. It's actually tighter than that, but it doesn't matter in the end because of the behavior of the denominator. Finally, subtracting one from both upper and lower bounds tells us us that


Thus the value of the numerator is essentially fixed between $-3$ and $1$. Finally, $\arctan{x}$ is also a bounded function in that its value for any input $x$ will never be smaller than $\frac{-\pi}{2}$ and never be larger than $\frac{\pi}{2}$. That means that our entire denominator $x+\arctan{x}$ will "act" like simply $x$. For extremely large values of $x$, adding $\frac{\pi}{2}$ is essentially negligible, since we are only considering large positive values of $x$.

Thus what we have is that

$$\frac{-3}{x+\arctan{x}}\le \frac{\sin{x}+\cos{x}-1}{x+\arctan{x}}\le\frac{1}{x+\arctan{x}}$$

Taking limits as $x\rightarrow \infty$,

$$\lim_{x\rightarrow\infty}\frac{-3}{x+\arctan{x}}\le \lim_{x\rightarrow\infty}\frac{\sin{x}+\cos{x}-1}{x+\arctan{x}}\le\lim_{x\rightarrow\infty}\frac{1}{x+\arctan{x}}$$

$$0\le \lim_{x\rightarrow\infty}\frac{\sin{x}+\cos{x}-1}{x+\arctan{x}}\le0$$

This forces our limit to be $0$ and means we have a horizontal asymptote $y=0$

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