Limits

99. Limits

99.0.1. Properties of Limits

99.0.1.1. Continuous

99.0.1.1.1. Addition, Subtraction, Multiplication, Division

\begin{matrix} lim_{𝑥 \to 𝑐} (𝑓 (𝑥) * 𝑔 (𝑥)) = lim_{𝑥 \to 𝑐} 𝑓 (𝑥) * lim_{𝑥 \to 𝑐} 𝑔 (𝑥) \\ * \in {+, -, \times, \div} \end{matrix}

99.0.1.1.2. Constant

lim_{𝑥 \to 𝑐} 𝑘 𝑓 (𝑥) = 𝑘 lim_{𝑥 \to 𝑐} 𝑓 (𝑥)

99.0.1.2. Non-continuous

Even though the limit for either function may not exist, their $*$ can exist as long as

\begin{matrix} lim_{𝑥 \to 𝑐^{-}} (𝑓 (𝑥) * 𝑔 (𝑥)) = lim_{𝑥 \to 𝑐^{+}} (𝑓 (𝑥) * 𝑔 (𝑥)) \\ * \in {+, -, \times, \div} \end{matrix}

Example

Problem 1: Limit Exists

lim_{𝑥 \to - 2} (𝑓 (𝑥) + 𝑔 (𝑥))

Left-hand limit ( $𝑥 \to - 2^{-}$ )

lim_{𝑥 \to - 2^{-}} 𝑓 (𝑥) = 1 lim_{𝑥 \to - 2^{-}} 𝑔 (𝑥) = 3

Adding these

\begin{aligned} lim_{𝑥 \to - 2^{-}} (𝑓 (𝑥) + 𝑔 (𝑥)) & = lim_{𝑥 \to - 2^{-}} 𝑓 (𝑥) + lim_{𝑥 \to - 2^{-}} 𝑔 (𝑥) \\ = 1 + 3 \\ = 4 \end{aligned}

Right-hand limit ( $𝑥 \to - 2^{+}$ )

lim_{𝑥 \to - 2^{+}} 𝑓 (𝑥) = 3 lim_{𝑥 \to - 2^{+}} 𝑔 (𝑥) = 1

Adding these

\begin{aligned} lim_{𝑥 \to - 2^{+}} (𝑓 (𝑥) + 𝑔 (𝑥)) & = lim_{𝑥 \to - 2^{+}} 𝑓 (𝑥) + lim_{𝑥 \to - 2^{+}} 𝑔 (𝑥) \\ = 3 + 1 \\ = 4 \end{aligned}

Since both the left-hand and right-hand limits agree

lim_{𝑥 \to - 2} (𝑓 (𝑥) + 𝑔 (𝑥)) = 4

Problem 2: Limit Does Not Exist

lim_{𝑥 \to 1} (𝑓 (𝑥) + 𝑔 (𝑥))

Left-hand limit ( $𝑥 \to 1^{-}$ )

lim_{𝑥 \to 1^{-}} 𝑓 (𝑥) = 2 lim_{𝑥 \to 1^{-}} 𝑔 (𝑥) = 0

Adding these

\begin{aligned} lim_{𝑥 \to 1^{-}} (𝑓 (𝑥) + 𝑔 (𝑥)) & = lim_{𝑥 \to 1^{-}} 𝑓 (𝑥) + lim_{𝑥 \to 1^{-}} 𝑔 (𝑥) \\ = 2 + 0 \\ = 2 \end{aligned}

Right-hand limit ( $𝑥 \to 1^{+}$ )

lim_{𝑥 \to 1^{+}} 𝑓 (𝑥) = - 1 lim_{𝑥 \to 1^{+}} 𝑔 (𝑥) = 0

Adding these

\begin{aligned} lim_{𝑥 \to 1^{+}} (𝑓 (𝑥) + 𝑔 (𝑥)) & = lim_{𝑥 \to 1^{+}} 𝑓 (𝑥) + lim_{𝑥 \to 1^{+}} 𝑔 (𝑥) \\ = - 1 + 0 \\ = - 1 \end{aligned}

Since both the left-hand and right-hand limits do not agree, the limit ${lim}_{𝑥 \to 𝑐} (𝑓 (𝑥) + 𝑔 (𝑥))$ does not exist

99.0.1.3. Composite Functions

lim_{𝑥 \to 𝑐} 𝑓 (𝑔 (𝑥)) = 𝑓 (lim_{𝑥 \to 𝑐} 𝑔 (𝑥))

For this to hold true, two important conditions must be satisfied:

Inner limit exists: The limit ${lim}_{𝑥 \to 𝑐}$ must exist and equal some value $𝐿$ . That is, as $𝑥$ gets arbitrarily close to $𝑐$ , $𝑔 (𝑥)$ approaches a well-defined number $𝐿$
Continuity of the outer function: The function $𝑓$ must be continuous at the point $𝐿$ . Continuity ensures that $𝑓$ behaves predictably near $𝐿$ , without any jumps, gaps, or undefined points

Example

Problem 1: Inner Limit & Continuity Exist

lim_{𝑥 \to - 3} 𝑓 (𝑔 (𝑥))

Inner limit ${lim}_{𝑥 \to - 3} 𝑔 (𝑥)$

Observing $𝑔 (𝑥)$ , as $𝑥 \to - 3$ , $𝑔 (𝑥) \to 3$ . The inner limit $𝐿 = 3$ exists

lim_{𝑥 \to - 3} 𝑔 (𝑥) = 3

Continuity of $𝑓 (𝑥)$ at $𝑥 = 3$

Observing $𝑓 (𝑥)$ , $𝑓 (3) = - 1$ . Since $𝑓 (𝑥)$ is continuous at $𝑥 = 3$ , the composite limit holds

\begin{aligned} 𝑓 (lim_{𝑥 \to - 3} 𝑔 (𝑥)) & = 𝑓 (3) \\ = - 1 \end{aligned}

Problem 2: Inner Limit Does Not Exist

lim_{𝑥 \to 2} 𝑓 (𝑔 (𝑥))

Inner limit ${lim}_{𝑥 \to 2} 𝑔 (𝑥)$

Observing $𝑔 (𝑥)$ , as $𝑥 \to 2$ , $𝑔 (𝑥) \to 2$ . The inner limit does not exist

Problem 3: Continuity Does Not Exist

lim_{𝑥 \to 0.5} 𝑓 (𝑔 (𝑥))

Inner limit ${lim}_{𝑥 \to 0.5} 𝑔 (𝑥)$

Observing $𝑔 (𝑥)$ , as $𝑥 \to 0.5$ , $𝑔 (𝑥) \to 1$ . The inner limit $𝐿 = 1$ exists

lim_{𝑥 \to 0.5} 𝑔 (𝑥) = 1

Continuity of $𝑓 (𝑥)$ at $𝑥 = 1$

Observing $𝑓 (𝑥)$ , $𝑓 (1)$ is not continuous. Since $𝑓 (𝑥)$ is not continuous at $𝑥 = 1$ , the composite limit does not hold

99.0.1.4. Limits by Direct Substitution

Example

Limit exists

\begin{aligned} lim_{𝑥 \to - 1} (6 𝑥^{2} + 5 𝑥 - 1) & = 6 {(- 1)}^{2} + 5 (- 1) - 1 \\ = 6 - 5 - 1 \\ = 0 \end{aligned}

Limit does not exist (Undefined)

\begin{aligned} lim_{𝑥 \to 1} \frac{𝑥}{ln (𝑥)} & = \frac{1}{ln (1)} \\ = \frac{1}{0} \end{aligned}

99.0.1.4.1. Limits of Piecewise Functions

99.0.1.4.2. Absolute Value

99.0.1.5. Limits by Factoring

99.0.1.6. Limits by Rationalizing

99.0.1.7. Continuity & Differentiability at a Point

Example

Piecewise function:

𝑓 (𝑥) = {\begin{cases} 𝑥^{2} & if 𝑥 < 3 \\ 6 𝑥 - 9 & if 𝑥 \geq 3 \end{cases}

Check for Continuity

Value of $𝑓 (3)$

𝑓 (3) = 6 (3) - 9 = 9

Left-Hand Limit (LHL)

lim_{𝑥 \to 3^{-}} 𝑓 (𝑥) = 3^{2} = 9

Right-Hand Limit (RHL)

lim_{𝑥 \to 3^{+}} 𝑓 (𝑥) = 6 (3) - 9 = 9

Since $𝑓 (3) = {lim}_{𝑥 \to 3^{-}} 𝑓 (𝑥) = {lim}_{𝑥 \to 3^{+}} 𝑓 (𝑥)$ , $𝑓 (𝑥)$ is continuous at $𝑥 = 3$

Check for Differentiability

Left-Hand Derivative (LHD)

\begin{aligned} lim_{𝑥 \to 3^{-}} \frac{𝑓 (𝑥) - 𝑓 (3)}{𝑥 - 3} & = \frac{𝑥^{2} - 3^{2}}{𝑥 - 3} \\ = \frac{𝑥^{2} - 9}{𝑥 - 3} \\ = \frac{(𝑥 + 3) (𝑥 - 3)}{𝑥 - 3} \\ = 𝑥 + 3 \\ = 6 \end{aligned}

Right-Hand Derivative

\begin{aligned} lim_{𝑥 \to 3^{+}} \frac{𝑓 (𝑥) - 𝑓 (3)}{𝑥 - 3} & = \frac{(6 𝑥 - 9) - 3^{2}}{𝑥 - 3} \\ = \frac{6 𝑥 - 9 - 9}{𝑥 - 3} \\ = \frac{6 𝑥 - 18}{𝑥 - 3} \\ = \frac{6 (𝑥 - 3)}{𝑥 - 3} \\ = 6 \end{aligned}

Since the left-hand and right-hand derivatives are equal, $𝑓 (𝑥)$ is differentiable at $𝑥 = 3$

Conclusion: $𝑓 (𝑥)$ is both continuous & differentiable at $𝑥 = 3$

Example