# What are the asymptote(s) and hole(s), if any, of  f(x) = xcos(1/x)?

Dec 13, 2015

$f \left(x\right)$ has a hole at $x = 0$ and a slant asymptote of y=x.

#### Explanation:

The function is undefined only at $x = 0$ (as there is a denominator of $x$). Then, to look for asymptotes and holes, we will check what happens as $x \to 0$, as well as the end behavior of the function, that is, what happens when $x \to \infty$ and when $x \to - \infty$.

First, looking as $x \to 0$:
Regardless of what happens to $\frac{1}{x}$, we have $- 1 \le \cos \left(\frac{1}{x}\right) \le 1$.
Thus $- x \le x \cos \left(\frac{1}{x}\right) \le x$, and so

as $x \to 0$, $x \cos \left(\frac{1}{x}\right) \to 0$.

Then there is no asymptote at $x = 0$, but $x \cos \left(\frac{1}{x}\right)$ is undefined at that point, meaning $f \left(x\right)$ has a hole at $x = 0$.

Now, looking at the end behavior, $\frac{1}{x} \to 0$ as $x \to \pm \infty$, meaning

$\cos \left(\frac{1}{x}\right) \to \cos \left(0\right) = 1$

Thus $x \cos \left(\frac{1}{x}\right) \to \pm \infty$ as $x \to \pm \infty$

So there are no horizontal asymptotes.
But $\cos \left(\frac{1}{x}\right)$ becomes closer and closer to $1$ as $x \to \pm \infty$, meaning $f \left(x\right)$ gets closer and closer to the line $y = x$. Thus $y = x$ is a slant asymptote for $f \left(x\right)$.