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.

# Left-Hand and Right-Hand Limits

 Definition: Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then the left-hand limit of $f$ at $c$ denoted $\lim_{x \to c^-} f(x) = L_1$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $x < c$ then $\mid f(x) - L_1 \mid < \epsilon$. Similarly, the right-hand limit of $f$ at $c$ denoted $\lim_{x \to c^+} f(x) = L_2$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $x > c$ then $\mid f(x) - L_2 \mid < \epsilon$.

By the definition of a left-hand limit, we are purely interested in what happens to the function to the left of the cluster point $c$, while for right-hand limits, we are purely interested in what happens to the function to the right of the cluster point $c$. There are five different cases that can happen with regards to left-hand and right-hand limits.

• Both the left-hand and right-hand limits exist and $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.
• Both the left-hand and right-hand limits exists but $\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$.
• Only the left-hand limit exists.
• Only the right-hand limit exists.
• Neither the left-hand limit nor the right-hand limit exists.

For example, consider the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \left\{\begin{matrix} x^2 & x > 1\\ \frac{3}{2} - x & x < 1 \end{matrix}\right.$. Notice that $\lim_{x \to 1^-} f(x) = \frac{1}{2}$ and $\lim_{x \to 1^+} f(x) = 1$, and so left-hand and right-hand limits need not be equal. We will now look at some theorems regarding left-hand and right-hand limits.

 Theorem 1 (Uniqueness of One-sided Limits): Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. If $\lim_{x \to c^-} f(x) = L_1$ and $\lim_{x \to c^-} f(x) = L_2$ then $L_1 = L_2$. Similarly if $\lim_{x \to c^+} f(x) = M_1$ and $\lim_{x \to c^+} f(x) = M_2$ then $M_1 = M_2$.

The proof of theorem 1 is analogous to that of The Uniqueness of Limits of a Function Theorem.

 Theorem 2: Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then $\lim_{x \to c} f(x) = L$ if and only if $\lim_{x \to c^-} f(x) = L = \lim_{x \to c^+} f(x)$.
• Proof: Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$.
• $\Rightarrow$ Suppose that $\lim_{x \to c} f(x) = L$. Then $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$.
• Therefore if $x < c$ and $x \in A$ are such that $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$ and so $\lim_{x \to c^-} f(x) = L$.
• Similarly if $x > c$ and $x \in A$ are such that $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$ and so $\lim_{x \to c^+} f(x) = L$.
• Therefore $\lim_{x \to c^-} f(x) = L = \lim_{x \to c^+} f(x)$.
• $\Leftarrow$ Now suppose that $\lim_{x \to c^-} f(x) = L = \lim_{x \to c^+}$. Notice that $x > c$ and $x < c$ is equivalent to saying $x \neq 0$, which is also equivalent to saying $0 < \mid x - c \mid$. Therefore $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$ and so $\lim_{x \to c} f(x) = L$. $\blacksquare$