why do we use limits approaches infinity to find horizontal asymptotes if it can't do some functions like sin(x)?

Question

I know that sin(x) has two horizontal asymptotes at ${y=1}$ and ${y=-1}$ but I can't prove it using this expression $$ \lim_{x \to \pm \infty} sin(x) = Indeterminate $$ so why finding horizontal asymptotes doesn't hold in this case?

Answer 1

It's important to consider what the definition of a "horizontal asymptote" is. A common misconception is that a horizontal asymptote of a function $f(x)$ is a horizontal line that $f(x)$ gets very close to but does not cross as $x$ tends to $\infty$; this is not, however, a good definition, because it wouldn't be a terribly useful notion -- that would mean that, for example, $\frac{x}{x^2 + 1}$ would have no horizontal asymptote!

The correct definition of a horizontal asymptote is a horizontal line which $f(x)$ gets and remains arbitrarily close to as $x$ tends to $\infty$ or $-\infty$. In other words, if you give me a small number $\varepsilon$, I can give you a big number $M$ so that whenever $x > M$, $f(x)$ is within a distance of $\varepsilon$ from that horizontal line. That's not true for $\sin{x}$, because if you give me $\varepsilon = 1/2$, for any $M$ I pick there will always be a point later on with $\sin{x}$ further than $1/2$ unit away from a maximum or a minimum. So $\sin{x}$ has no horizontal asymptotes -- which is precisely what the limit is telling you!

Answer 2

Because $y=\pm 1$ are not asymptotes of $\sin x$. An asymptote of a function $f:\mathbb R\rightarrow\mathbb R$ is a line $y=mx+q$ such that $$\lim_{x\rightarrow\infty}(f(x)-mx-q)=0$$ this also gives the formulas for $m,q$, that is $q=\lim_{x\rightarrow\infty}(f(x)-mx)$ and $m=\lim_{x\rightarrow\infty}\frac{f(x)}{x}$. (the $\infty$ is either $+\infty$ or $-\infty$, a function may have no asymptote, like $\sin x$, the same asymptote for $\pm\infty$, like $\frac{1}{x}$ or two different asymptote for $\pm\infty$, like $\arctan x$).

No such line exists for $\sin x$ (you can use the formulas above for $m,q$ and try to find the values for $f(x)=\sin x$, you'll find out that the limit for $q$ doesn't exists)