*This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.*

# Limit Divergence Criteria

Recall from The Sequential Criterion for a Limit of a Function page, that for a function $f : A \to \mathbb{R}$ and for $c$ being a cluster point of $A$, then $\lim_{x \to c} f(x) = L$ if and only if for all sequences $(a_n)$ from $A$ for which $\lim_{n \to \infty} a_n = c$ we also have that $\lim_{n \to \infty} f(a_n) = L$. We will now formulate what is known as the Limit Divergence Criteria, which will establish criteria to establish whether the value $L$ is not the limit of $f$ as $x \to c$.

Theorem 1 (Limit Divergence Criteria): Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then for $L \in \mathbb{R}$, $\lim_{x \to c} f(x) \neq L$ if either:1. There exists a sequence $(a_n)$ from $A$ where $a_n \neq c$ $\forall n \in \mathbb{N}$ such that $\lim_{n \to \infty} a_n = c$ but $\lim_{n \to \infty} f(a_n) \neq L$.2. There exists a sequence $(a_n)$ from $A$ where $a_n \neq c$ $\forall n \in \mathbb{N}$ such that $\lim_{n \to \infty} a_n = c$ but $(f(a_n))$ does not converge in $\mathbb{R}$. |

**Proof:**Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Let $L \in \mathbb{R}$.

- If either
**(1)**or**(2)**holds, then we note that this contradicts The Sequential Criterion for a Limit of a function. Since in both cases 1 and 2, not all sequences $(a_n)$ where $a_n \neq c$ $\forall n \in \mathbb{N}$ from $A$ such that $\lim_{n \to \infty} a_n = c$ satisfy $\lim_{n \to \infty} f(a_n) = L$, and so we have that $\lim_{x \to c} f(x) \neq L$. $\blacksquare$

The following diagram represents the first scenario of the Limit Divergence Criteria for which there exists a sequence, in this case $(b_n)$ that converges to $c$, however $(f(b_n))$ does not converge to $L$ and so $\lim_{x \to c} f(x) \neq L$:

The succeeding diagram represents the second scenario of the Limit Divergence Criteria for which there exists a sequence, $(b_n)$ that converges to $c$, however $(f(b_n))$ does not converge in $\mathbb{R}$ and so $\lim_{x \to c} f(x) \neq L$:

## Example 1

**Using the Divergence Criteria, for the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{1}{x - 1}$ show that $\lim_{x \to 1} \frac{1}{x - 1}$ does not exist in $\mathbb{R}$.**

Consider the sequence $(a_n) = \frac{1}{n} + 1$. We note that this sequence $(a_n)$ is from $\mathbb{R}$ (the domain). Also notice that this sequence converges to $1$ as $\lim_{n \to \infty} \frac{1}{n} + 1 = 1$. Now notice that $(f(a_n)) = \frac{1}{\frac{1}{n} + 1 - 1} = n$, and $\lim_{n \to \infty} f(a_n) = \lim_{n \to \infty} n = \infty$.

Therefore, we have a sequence $(a_n)$ from the domain that converges to $1$, however the sequence $(f(a_n))$ does not converge in $\mathbb{R}$. By **(2)** of the Divergence Criteria, we have that $\lim_{x \to 1} \frac{1}{x - 1}$ does not exist in $\mathbb{R}$.