The Sequential Criterion for Left-Hand and Right-Hand Limits

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.

# The Sequential Criterion for Left-Hand and Right-Hand Limits

Recall from the Left-Hand and Right-Hand Limits page that for a function $f : A \to \mathbb{R}$ where $c$ is a cluster point of $A$, then $\lim_{x \to c^-} f(x) = L_1$ means that $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$, $x < c$, and $0 < \mid x - c \mid < \delta$, then $\mid f(x) - L_1 \mid < \epsilon$, and we say that $L_1$ is the left-hand limit of $f$ at $c$.

Similarly $\lim_{x \to c^+} f(x) = L_2$ means that $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$, $x > c$, and $0 < \mid x - c \mid < \delta$, then $\mid f(x) - L_2 \mid < \epsilon$ and we say that $L_2$ is the right-hand limit of $f$ at $c$.

We will now look at what is known as The Sequential Criterion for Left-Hand and Right-Hand Limits, which is analogous to that of normal limits. For left-hand limits, it says that for any sequence $(a_n)$ of real numbers to the left of $c$ for which $(a_n)$ converges to $c$, then the sequence $(f(a_n))$ converges to $L_1$. For right-hand limits, it says that for any sequence $(b_n)$ of real numbers to the right of $c$ for which $(b_n)$ converges to $c$, then the sequence $(f(a_n))$ converges to $L_2$. We will summarize this in the following theorem.

 Theorem 1 (Sequential Criterion, Left-Hand Limits): Let $f : A \to \mathbb{R}$ be a function where $c$ is a cluster point of $A$. Then if $\lim_{x \to c^-} f(x) = L_1$ then for every sequence $(a_n)$ for which $a_n \in A$ and $a_n < c$ $\forall n \in \mathbb{N}$ that converges to $c$, then $(f(a_n))$ converges to $L_1$. • Proof: $\Rightarrow$ Suppose that $\lim_{x \to c^-} f(x) = L_1$ and let $(a_n)$ be a sequence in $A$ such that $a_n < c$ $\forall n \in \mathbb{N}$ and such that $\lim_{n \to \infty} a_n = c$. We want to then show that $\lim_{n \to \infty} f(a_n) = L_1$.
• Let $\epsilon > 0$ be given. Since $\lim_{x \to c^-} f(x) = L_1$ then $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $x < c$ and $\mid x - c \mid < \delta$ then $\mid f(x) - L_1 \mid < \epsilon$. Now since $\delta > 0$, since we have that $\lim_{n \to \infty} a_n = c$ and $a_n < c$ $\forall n \in \mathbb{N}$, then for $\delta > 0$ $\exists N \in \mathbb{N}$ such that if $n \geq N$ then $\mid a_n - c \mid < \delta$. Therefore $\forall n \geq N$ it must be that $\mid f(a_n) - L_1 \mid < \epsilon$ and so $\lim_{x \to c^-} f(x) = L_1$.
• $\Leftarrow$ Suppose that for all sequences $(a_n)$ in $A$ such that $a_n < c$ then if $\lim_{n \to \infty} a_n = c$ then $\lim_{n \to \infty} f(a_n) = L_1$. We want to then show that $\lim_{x \to c^-} f(x) = L_2$.
• Suppose not, in other words suppose that $\exists \epsilon_0 > 0$ such that $\forall \delta > 0$ then $\exists x_{\delta} \in A$, $x_{\delta} < c$ such that $0 < \mid x_{\delta} - c \mid < \delta$ however $\mid f(x_{\delta}) - L_1 \mid \geq L_1$.
• Consider $\epsilon_0 > 0$ and let $\delta_n = \frac{1}{n}$. Then since $\lim_{n \to \infty} a_n = c$ then for $\delta_n = \frac{1}{n} > 0$ then $\exists N \in \mathbb{N}$ such that if $n \geq N$ then $\mid a_n - c \mid < \delta_n = \frac{1}{n}$. However this implies that $\mid f(a_n) - L_1 \mid \geq \epsilon_0$ and so $\lim_{n \to \infty} f(a_n) \neq L_1$, a contradiction. Therefore our assumption that $\lim_{x \to c^-} f(x) \neq L_1$ was false, and so $\lim_{x \to c^-} f(x) = L_1$. $\blacksquare$
 Theorem 2 (Sequential Criterion, Right-Hand Limits): Let $f : A \to \mathbb{R}$ be a function where $c$ is a cluster point of $A$. Then if $\lim_{x \to c^+} f(x) = L_2$ then for every sequence $(a_n)$ for which $a_n \in A$ and $a_n > c$ $\forall n \in \mathbb{N}$ that converges to $c$, then $(f(a_n))$ converges to $L_2$. An analogous proof can be obtained for The Sequential Criterion for Right-Hand Limits from above.