What is $\lim_{x \to \infty} \frac{\sin x}{\sin x}?$

As the title suggests, I am interested in computing $$\lim_{x \to \infty} \frac{\sin x}{\sin x}.$$ Some of my friends say that the limit is $1$ because the domain of function is $\mathbb{R} \setminus \{n \pi : n \text{ is an integer}\},$ whereas others say that this limit is undefined.

Which opinion is correct?

Solution 1:

It depends on exactly what your definition of limit is.

In calculus, limits of real-valued functions are often defined in such a way that in order for the limit as $x\to a$ to exist, the function must be defined on some interval of the form $(a-\delta,a+\delta)$, except perhaps at $a$. For limits as $x\to\infty$, you require the function to be defined on some interval of the form $(M,\infty)$. The definition using $\epsilon$s would look something like:

$\lim_{x\to\infty} f(x)=L$ if and only if for all $\epsilon\gt 0$ there exists $M$ such that if $x\gt M$, then $|f(x)-L|\lt\epsilon$.

However, sometimes limits are defined differently (slightly more generally, applicable to more functions), and the prerequisite for the limit as $x\to a$ to exist is that $a$ be an “accumulation point” of the domain of $f$. That is, that for all $\delta\gt 0$, there exist points $b$ in the domain of $f$ such that $0\lt |b-a|\lt \delta$. For limits as $x\to\infty$, one requires that for all $M$ there exist $b$ in the domain of $f$ such that $b\gt M$.

Here, the definition using $\epsilon$s would be:

$\lim_{x\to\infty} f(x)=L$ if and only if (i) for every $M$ there exists $x\gt M$ such that $x\in\mathrm{dom}(f)$; and (ii) for every $\epsilon\gt 0$ there exists $N$ such that if $x\gt N$ and $x\in\mathrm{dom}(f)$, then $|f(x)-L|\lt\epsilon$.

If your definition of limits is of the first type, then the limit you write does not exist, because the function does not meet the requirement to be able to even talk about a potential limit as $x\to\infty$: it is not defined on any interval of the form $(M,\infty)$, so you can always find an $x$ with $x\gt M$, and where $|f(x)-1|\lt\epsilon$ fails because the expression doesn’t even make sense. On the other hand, if your definition of limit is of the second type, then the limit is $1$, since the satisfies both conditions of that definition.

In short, neither opinion is right and neither is wrong: it depends on exactly what the symbol $\lim_{x\to\infty}f(x)=L$ means.

Solution 2:

By the familiar definition (i.e., the "Wikipedia" definition), for a real-valued function $f(x),$ we have that $\lim_{x \to \infty} f(x)$ is a real number $L$ (if it exists) such that for every real number $\varepsilon > 0,$ there exists a real number $M$ such that for all $x > M,$ we have that $|f(x) - L| < \varepsilon.$

Like you have mentioned, if $x$ is not an integer multiple of $\pi,$ we have that $f(x) = \frac{\sin x}{\sin x} = 1$; however, for any integer $n,$ we have that $f(n \pi)$ is not defined. What can you conclude?