Does $\lim_{x \to 1}\left(\frac{x}{[x]}\right)$ exist?

Question

Does the limit $\lim_{x \to 1}\left(\frac{x}{\lfloor x\rfloor}\right)$ exist? Where $\lfloor x\rfloor$ is the Greatest Integer Function or the Floor function.

My teacher defined that a limit exists when the Left-hand Limit = Right-Hand Limit and If any one is undefined but the other is finite then also the limit exists. Like $\lim_{x \to 0}\sqrt{x}$ does exist although the LHL does not lie in the domain.

In this question the domain of the function $\frac{x}{\lfloor x\rfloor}$ is $\mathbb R\setminus[0,1)$ where $\mathbb R$ represents Real Numbers. So, The Left-Hand Limit is not defined and Right Hand Limit is $1$, which is finite so according to my teacher Limit does exist. But when I tried this question in WolframAlpha, it is saying that the LHL is $-\infty$ and RHL is $1$, so limit does not exist.

So, a more general question is: do we have to consider the domain of a function when we calculate a limit?

Answer 1

Your teacher is completely right, according to the more general definition of limit, we consider $x$ in the domain and therefore

$$\lim_{x \to 1}\left(\frac{x}{[x]}\right)= \lim_{x \to 1^+}\left(\frac{x}{[x]}\right)=1$$

Refer also to the related

• Calculating $\lim_{x \to 0}{\frac{ \lfloor x \rfloor}{\lfloor x \rfloor}}$
• What is $\lim_{x \to 0}\frac{\sin\left(\frac 1x\right)}{\sin \left(\frac 1 x\right)}$ ? Does it exist?

Answer 2

First of all to solve this question of limit first of all you have to be familiar with the definition of limit.

$$\lim_{x\to a} f(x)$$ will exist only when both the left hand and right hand limits will exist. That is $$\lim_{x\to a} f(x)$$ will exist only when $$\lim_{x\to a^{+}}f(x)$$ and $$\lim_{x\to a^{-}}f(x)$$ both will exist.

Now according to your question $$\lim_{x\to 1}(\frac{x}{[x]})$$ will exist only when $$\lim_{x\to 1^{+}}(\frac{x}{[x]})$$ and $$\lim_{x\to 1^{-}}(\frac{x}{[x]})$$ both will exist. Now the interesting fact is that $$\lim_{x\to 1^{+}} [x] = 1$$ and $$\lim_{x\to 1^{-}} [x] = 0$$. So, as far as I can guess $$\lim_{x\to 1^{-}}(\frac{x}{[x]})$$ will be undefined. Therefore since $$\lim_{x\to 1^{-}}(\frac{x}{[x]})$$ is undefined so, $$\lim_{x\to 1}(\frac{x}{[x]})$$ will not exist.

Though I am not sure whether my logic is correct or not. But I would suggest to verify once.