Left-Hand and Right-Hand Limits

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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

\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

\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.

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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 $\lim_{x \to c} f(x) = L$ . $\blacksquare$

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Left-Hand and Right-Hand Limits