# How to prove a limit of a function does not exist?

I am looking at the function $$f(x) = \begin{cases} \dfrac{x^2-1}{x^2-x} & x \ne 0,1\\ 0 & x=0\\ 2 &x=1 \end{cases}$$ and am trying to show that $$\lim_{x \to 0} f(x)$$ DNE. This makes sense to me because $$f$$ goes towards $$+\infty$$ from the left and $$-\infty$$ from the right. However, in my analysis class we do not have a definition for when a limit does not exist. My first instinct was to negate the definition of when a limit exists, but this means that the limit would not exist for any number I chose, and I want to prove that the limit doesn't exist for any numbers.

I don't understand why there needs to be cases for this type of problem as detailed here.

How can I go about proving this rigorously?

• Your intuition is correct. Show that the limits differ when $x$ is approaching zero from the left and from the right. If $\lim \limits_{x \to 0} g(x)$ exists, these two limits must be the same. Feb 28, 2022 at 19:52
• You wrote "in my analysis class we do not have a definition for when a limit does not exist". But your instinct is correct: once you have any definition for when something does exist, you simply negate that statement to get the definition for when that thing does not exist. Apr 22, 2022 at 1:45
• If you assume that the limit is a real number $L$, without specifying any particular number, and then show the negation of the definition that the limit is $L$, you have just shown by contradiction that $L$ is not the limit, and since you didn't say which real number $L$ was, you've also shown that the limit doesn't exist for any number. Apr 22, 2022 at 2:05

Here is the formal proof based on $$\epsilon-\delta.$$

Note $$f(x)=\begin{cases}1+\dfrac{1}{x} \ &\mathrm{if} \ x\neq 0\\ 0 \ &\mathrm{if}\ x=0 \end{cases}$$.

Suppose $$\lim_{x\to 0}f(x)$$ exists. Let $$\lim_{x\to 0}f(x)=\alpha.$$

Then, there exists $$\delta\in(0,1)$$ s.t. $$0<|x|<\delta \Rightarrow |f(x)-\alpha|<2$$.

From this, $$|f(\frac{\delta}{2})-\alpha|<2$$ and $$|f(-\frac{\delta}{2})-\alpha|<2$$ must hold.

From $$|f(\frac{\delta}{2})-\alpha|<2$$, we get $$-2<1+\frac{2}{\delta}-\alpha<2$$ ・・・(i)

From $$|f(-\frac{\delta}{2})-\alpha|<2$$, we get $$-2<1-\frac{2}{\delta}-\alpha<2$$, i.e., $$-2<-1+\frac{2}{\delta}+\alpha<2$$ ・・・(ii)

Consider (i)+(ii). We get $$-4<\frac{4}{\delta}<4$$, thus $$1<\delta.$$

However, this contradicts $$\delta\in(0,1)$$.

Thus, the supposition "$$\lim_{x\to 0}f(x)$$ exists" is false.

Therefore, $$\lim_{x\to 0} f(x)$$ doesn't exist.

Drawing a picture helps. With a picture, you observe that the graph approaches $$-\infty$$ from the left and it approaches $$\infty$$ from the right. This is because $$\frac{x^2-1}{x^2-x}=\frac{x+1}{x}=1+\frac{1}{x}$$ for $$x \neq 1$$ (the parent function of hyperbolas shifted up by $$1$$ unit), and all we care about is what happens near $$x=0$$.