algebra precalculus - Asymptote to $frac{sin x}{x}$?

I have seen elsewhere that:

$$y=\frac{\sin x}{x}$$

has a horizontal asymptote of $y=0$, as it approaches that line as $x$ tends to $\pm \infty$.

Now, why does it not have an asymptote of $x=0$ or $y=1$, as the curve tends towards but never touches these lines? (Which satisfies the definition given by wolfram alpha)

Answer

Let's make our definition of an asymptote more clear.

An vertical asymptote for $x=a$ occurs if $\lim_{x\to a^+}f(x)=\pm\infty$ and $\lim_{x\to a^-}f(x)=\pm\infty$. The limit from the left does not have to equal the limit on the right, in fact there is an asymptote as long as one side goes to $\pm\infty$. Take the asymptote of $f(x)=\ln(x)$ for example.

A horizontal asymptote for $y=b$ occurs if $\lim_{x\to\infty}f(x)=b$ or if $\lim_{x\to-\infty}f(x)=b$, where $b$ is finite.

We can also have a curved asymptote. Say $f(x)$ is asymptotic to $g(x)$, then $\lim_{x\to\pm\infty}f(x)-g(x)=0$. The two functions must both get really close to each other as $x$ becomes arbitrarily large.

So to answer your questions, $y=0$ is an asymptote but $x=0$ and $y=1$ are not.