Limits paradox in case of sine function

Question

What according to you will be the answer for: $$\lim_{x \to 0} \frac{\sin(1/x)}{\sin(1/x)}$$

My answer was $1$. The book says not defined and the argument is that both numerator and denominator are in $[-1,1]$, so we cannot determine what it exactly is.

My argument is, whatever value it takes, shouldn't it be the same in both numerator and denominator? Moreover, another example may be:

$$\lim_{x \to 0} \frac{1/x}{1/x}$$

I say $1$ again, because it is the same. But, it is not defined if we individually consider numerator and denominator! Still we have cancelled it in this case. Why can't I do it in case of sine function?

Answer 1

The limit does not exist, but not because "both numerator and denominator are in [-1,1], so we can not determine what it exactly is". The denominator is $0$ whenever $\frac{1}{x}$ is a multiple of $π$, so in any open interval containing $0$ there is a point at which the expression is not defined, and hence the limit does not exist. This has nothing to do with the oscillatory behaviour of the numerator and denominator.

In your other example, the limit is indeed $1$ because it is indeed defined and equal to $1$ in any open interval of $0$ less $0$ itself.

Answer 2

When considering $\lim_{x\to 0} f(x)$, it does not matter whether or not $f(0)$ is defined at all. However, we still should have a look at the domain where your function $f(x)=\frac{\sin(1/x)}{\sin(1/x)}$ is defined. The problem is that we may sometimes divide by $0$, namely first of all $\frac1x$ is not defined for $x=0$, but also $\sin(1/x)=0$ whenever $\frac1x$ is an integer multiple of $\pi$. Hence the maximal domain of $f$ is $$D=\mathbb R\setminus\bigl(\{0\}\cup\{\,\tfrac1{k\pi}\mid k\in\mathbb Z\setminus\{0\}\,\}\bigr).$$ We note that for arbitrary $\epsilon>0$, the set $D\cap(-\epsilon,\epsilon)$ has more gaps than just at $x=0$ and I can only imagine that this is what your book objects to. On the other hand, it makes sense to define $\lim_{x\to a} f(x)$ for all $a\in\overline D$ and that would include tha case $a=0$. Only an "unlucky" wording of the definition of limit may prevent this, so you should check back what the exact definition of $\lim_{x\to a}f(x)$ is in your book (including, what conditions are imposed about where $f$ is defined).

Answer 3

Go back to basics: if you would like to show the limit is 1 as x approaches 0, then given $ \epsilon > 0$ can you find $\delta > 0$ such that $|f(x) - 1| < \epsilon$ for ALL $0<|x| < \delta$ ?

The answer is NO because there are an (infinite) number of points in $(-\delta, 0) \cup (0,+\delta)$ where $f(x)$ is undefined - i.e. all those points as noted in previous answers where $1/x$ is a multiple of $\pi$.

Answer 4

I don't know if it should be a comment or an answer:

In most books, the definition of limit is: given $ \epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ for all $0<|x| < \delta$. Using this definition, there is no limit, as explained by many other answers.

However, some books use: given $ \epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ for all $x$ in the domain of $f$ satisfying $0<|x| < \delta$. With this second definition, then your function is constant on its domain of definition, and the limit exists.

I guess your book is using the first definition.

Answer 5

My view of the problem: you function $f$ is defined on $D=(0,1]\setminus\{x\in [0,1]:\frac{1}{x\pi}\in\Bbb N\}$. Easy to see that $f\equiv 1$ on $D$. Therefore we can, by continuity, say that $f=1$ on $\bar D=[0,1]$.

So, in other words, the limit does not exist as it is, because in each neighbourhood of $0$ there's a point where the function is not defined explicitly. However, in all of those points we can define our function by continuity and then pass to the limit $x\to 0$.