Why can $y=0$ be considered the asymptote of $f(x)=\frac{\sin x}{x}$?

Question

Why can $y=0$ be considered the asymptote of $f(x)=\frac{\sin x}{x}$ when the two graphs don't get any closer when $x$ approaches infinity? Because isn't the asymptote something that a graph will touch and stay on once the graph reaches infinity? And yet, in $f(x)=\frac{\sin x}{x}$, the size of $x$ has no effect on the graph's proximity to the asymptote. It just keeps fluctuating around $y=0$! So that suggests that the graph won't touch and stay on the asymptote at infinity? And if it doesn't, how can $y=0$ still be considered the asymptote of $f(x)=\frac{\sin x}{x}$?

Answer 1

The definition of an horizontal asymptote is that $$\lim\limits_{x \to \infty} f(x) = c.$$

Which is exactly the case here with $c=0$.

In no way the definition of an asymptote implies that the function is supposed to stay on the same side of the asymptote for $x$ large enough.

Answer 2

f(x)=sinx/x, the size of x has no effect on the graph's proximity to the asymptote.

Yes it does: $$\text{max}_{x>x_0}\left|\frac{\sin(x)}{x}-0\right|\leq \frac{1}{x_0}$$

Answer 3

I have already written to you about this. Look at the graph of your function:

Don't you agree that, as we move to the right, the graph gets closer to $0$? To be more precise, if we fix a number $\varepsilon>0$, then, if $M$ is large enough, the graph of the restriction of $f$ to $[M,+\infty)$ is between the lines $y=\varepsilon$ and $y=-\varepsilon$. That's what it means that the line $y=0$ is an asymptote of the graph of $f$.

Answer 4

There are several misconceptions in your post.

$$\frac{\sin x}{x}$$ does get closer and closer to $0$ when $x$ increases. In fact,

$$\left|\frac{\sin x}{x}\right|\le\frac1x$$ and the RHS clearly tends to $0$.

"... isn't the asymptote something that a graph will touch and stay on ...": no, quite often an asymptote doesn't touch the curve. It just needs to come arbitrarily close. In the case at hand, it crosses it infinitely often, but this does not matter.