Limits and Continuity Complete Cheatsheet | Formulas & Examples

Quick Reference - Key Formulas

Definition of Limits

Intuitive Definition: The limit of $f(x)$ as $x$ approaches $c$ is $L$ if the values of $f(x)$ get arbitrarily close to $L$ as $x$ gets arbitrarily close to $c$.

ε-δ Definition (Formal): For every $\epsilon > 0$, there exists a $\delta > 0$ such that:

Example: Find $\lim_{x \to 2} (3x + 1)$.

Solution: As $x$ approaches 2, $3x + 1$ approaches $3(2) + 1 = 7$. Therefore, $\lim_{x \to 2} (3x + 1) = 7$.

Properties of Limits

If $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$, then:

Sum/Difference Rule:

Example: $\lim_{x \to 1} [(x^2 + 2x) + (3x - 1)] = \lim_{x \to 1} (x^2 + 2x) + \lim_{x \to 1} (3x - 1) = (1 + 2) + (3 - 1) = 7$

Product Rule:

Example: $\lim_{x \to 2} [x \cdot (x + 1)] = \lim_{x \to 2} x \cdot \lim_{x \to 2} (x + 1) = 2 \cdot 3 = 6$

Quotient Rule:

Example: $\lim_{x \to 3} \frac{x^2}{x + 1} = \frac{\lim_{x \to 3} x^2}{\lim_{x \to 3} (x + 1)} = \frac{9}{4}$

Power Rule:

Example: $\lim_{x \to 2} (x + 1)^3 = [\lim_{x \to 2} (x + 1)]^3 = 3^3 = 27$

Constant Multiple Rule:

Example: $\lim_{x \to 1} [5x^2] = 5 \cdot \lim_{x \to 1} x^2 = 5 \cdot 1 = 5$

One-Sided Limits

Left-Hand Limit: $\lim_{x \to c^-} f(x)$ - The limit as $x$ approaches $c$ from the left (values less than $c$).

Right-Hand Limit: $\lim_{x \to c^+} f(x)$ - The limit as $x$ approaches $c$ from the right (values greater than $c$).

Important Result: $\lim_{x \to c} f(x) = L$ exists if and only if:

Example: Find the one-sided limits of $f(x) = \frac{|x|}{x}$ at $x = 0$.

Solution:

• Left-hand limit: $\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = \lim_{x \to 0^-} (-1) = -1$
• Right-hand limit: $\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} 1 = 1$

Since the left and right limits are different, $\lim_{x \to 0} f(x)$ does not exist.

Limits at Infinity

Horizontal Asymptotes: Describe the behavior of functions as $x \to \pm\infty$.

For Rational Functions: If $f(x) = \frac{a_n x^n + \ldots + a_0}{b_m x^m + \ldots + b_0}$:

• If $n < m$: $\lim_{x \to \infty} f(x) = 0$
• If $n = m$: $\lim_{x \to \infty} f(x) = \frac{a_n}{b_m}$
• If $n > m$: $\lim_{x \to \infty} f(x) = \pm\infty$

Example 1: Find $\lim_{x \to \infty} \frac{3x^2 + 2x + 1}{x^2 - 5}$.

Solution: Since the degrees are equal (both 2), the limit is $\frac{3}{1} = 3$.

Example 2: Find $\lim_{x \to \infty} \frac{2x + 1}{x^2 + 3}$.

Solution: Since the degree of numerator (1) is less than denominator (2), the limit is $0$.

Infinite Limits: $\lim_{x \to c} f(x) = \pm\infty$ indicates vertical asymptotes.

Example: $\lim_{x \to 0^+} \frac{1}{x} = +\infty$ and $\lim_{x \to 0^-} \frac{1}{x} = -\infty$

Standard Limits

These are fundamental limits that appear frequently in calculus:

Trigonometric Limits:

• $\lim_{x \to 0} \frac{\sin x}{x} = 1$
• $\lim_{x \to 0} \frac{\tan x}{x} = 1$
• $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$

Example: Find $\lim_{x \to 0} \frac{\sin 3x}{x}$.

Solution: $\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot 3 = 1 \cdot 3 = 3$

Exponential and Logarithmic Limits:

• $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$
• $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
• $\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1$
• $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$

Example: Find $\lim_{x \to 0} \frac{e^{2x} - 1}{x}$.

Solution: $\lim_{x \to 0} \frac{e^{2x} - 1}{x} = \lim_{x \to 0} \frac{e^{2x} - 1}{2x} \cdot 2 = 1 \cdot 2 = 2$

Indeterminate Forms and L'Hôpital's Rule

Indeterminate Forms: $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$, $\infty^0$

L'Hôpital's Rule: If $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$ (or both approach $\pm\infty$), then:

(provided the right-hand limit exists)

Example 1: Find $\lim_{x \to 0} \frac{x^2}{1 - \cos x}$.

Solution: This is $\frac{0}{0}$ form. Applying L'Hôpital's rule:

Example 2: Find $\lim_{x \to \infty} \frac{x^2}{e^x}$.

Solution: This is $\frac{\infty}{\infty}$ form. Applying L'Hôpital's rule twice:

Algebraic Method Example: Find $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$.

Solution: Factor the numerator: $\lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2$

Squeeze Theorem (Sandwich Theorem)

Statement: If $g(x) \leq f(x) \leq h(x)$ for all $x$ in some interval containing $c$ (except possibly at $c$), and if:

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

Example: Find $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$.

Solution: Since $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$, we have:

Since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$, by the Squeeze Theorem:

Definition of Continuity

A function $f(x)$ is continuous at $x = c$ if:

1. $f(c)$ exists (function is defined at $c$)
2. $\lim_{x \to c} f(x)$ exists
3. $\lim_{x \to c} f(x) = f(c)$

Equivalently: $f$ is continuous at $c$ if:

Example: Check if $f(x) = \begin{cases} x^2 & \text{if } x \neq 2 \\ 5 & \text{if } x = 2 \end{cases}$ is continuous at $x = 2$.

Solution:

• $f(2) = 5$ ✓ (function is defined)
• $\lim_{x \to 2} f(x) = \lim_{x \to 2} x^2 = 4$ ✓ (limit exists)
• $\lim_{x \to 2} f(x) = 4 \neq 5 = f(2)$ ✗

Since the limit doesn't equal the function value, $f$ is not continuous at $x = 2$.

Types of Discontinuities

1. Removable Discontinuity (Point Discontinuity):

Occurs when $\lim_{x \to c} f(x)$ exists but either $f(c)$ doesn't exist or $\lim_{x \to c} f(x) \neq f(c)$.

Example: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.

Here, $\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4$, but $f(2)$ is undefined.

2. Jump Discontinuity:

Occurs when $\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$ (one-sided limits exist but are different).

Example: $f(x) = \begin{cases} x + 1 & \text{if } x < 1 \\ x + 2 & \text{if } x \geq 1 \end{cases}$ at $x = 1$.

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

3. Infinite Discontinuity:

Occurs when $\lim_{x \to c} f(x) = \pm\infty$.

Example: $f(x) = \frac{1}{x-3}$ at $x = 3$.

$\lim_{x \to 3^-} f(x) = -\infty$ and $\lim_{x \to 3^+} f(x) = +\infty$

4. Oscillatory Discontinuity:

Occurs when the function oscillates infinitely as it approaches the point.

Example: $f(x) = \sin\left(\frac{1}{x}\right)$ at $x = 0$.

Continuity on Intervals

Continuous on Open Interval (a, b): $f$ is continuous at every point in $(a, b)$.

Continuous on Closed Interval [a, b]: $f$ is continuous on $(a, b)$ and:

• $\lim_{x \to a^+} f(x) = f(a)$ (right-continuous at $a$)
• $\lim_{x \to b^-} f(x) = f(b)$ (left-continuous at $b$)

Properties of Continuous Functions:

• Polynomial functions are continuous everywhere
• Rational functions are continuous wherever the denominator is non-zero
• Trigonometric functions are continuous on their domains
• Exponential and logarithmic functions are continuous on their domains
• Sum, difference, product, and quotient of continuous functions are continuous (quotient where denominator ≠ 0)
• Composition of continuous functions is continuous

Important Theorems

Intermediate Value Theorem (IVT):

If $f$ is continuous on $[a, b]$ and $k$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c \in [a, b]$ such that $f(c) = k$.

Application: Proving existence of roots of equations.

Example: Show that $f(x) = x^3 - x - 1$ has a root in $[1, 2]$.

Solution: $f(1) = -1 < 0$ and $f(2) = 5 > 0$. Since $f$ is continuous and $0$ lies between $-1$ and $5$, by IVT, there exists $c \in [1, 2]$ such that $f(c) = 0$.

Extreme Value Theorem:

If $f$ is continuous on $[a, b]$, then $f$ attains its maximum and minimum values on $[a, b]$.

Uniform Continuity:

A function $f$ is uniformly continuous on an interval if for every $\epsilon > 0$, there exists $\delta > 0$ such that for all $x, y$ in the interval:

Limits and Continuity of Piecewise Functions

Example: Analyze the continuity of $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 0 \\ 2x + 1 & \text{if } x \geq 0 \end{cases}$ at $x = 0$.

Solution:

• $f(0) = 2(0) + 1 = 1$
• $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^2 + 1) = 1$
• $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x + 1) = 1$

Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1$, the function is continuous at $x = 0$.

General Approach for Piecewise Functions:

1. Check continuity at boundary points
2. Verify that left and right limits are equal
3. Ensure the function value equals the limit
4. Each piece is continuous on its domain

Practical Problem-Solving Techniques

For Limit Problems:

1. Direct Substitution: Try substituting the value first
2. Factoring: Factor and cancel common terms
3. Rationalization: Multiply by conjugate for radical expressions
4. L'Hôpital's Rule: For indeterminate forms
5. Standard Limits: Use known limit formulas
6. Squeeze Theorem: For oscillating functions

Rationalization Example: Find $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$.

Solution: Multiply by conjugate:

For Continuity Problems:

1. Check if the function is defined at the point
2. Calculate the limit at the point
3. Compare the limit with the function value
4. For piecewise functions, check boundary points carefully

Advanced Topics

Limits of Sequences:

For sequences $\{a_n\}$, we write $\lim_{n \to \infty} a_n = L$ if the terms get arbitrarily close to $L$ as $n$ increases.

Example: $\lim_{n \to \infty} \frac{1}{n} = 0$, $\lim_{n \to \infty} \frac{n+1}{n} = 1$

Continuity and Differentiability:

If a function is differentiable at a point, then it is continuous at that point. However, continuity does not imply differentiability.

Example: $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there.

Limits of Composite Functions:

If $\lim_{x \to c} g(x) = L$ and $f$ is continuous at $L$, then:

Common Mistakes to Avoid

Practice Problems

Problem 1: Find $\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 6x + 9}$.

Solution: Factor both numerator and denominator:

As $x \to 3$, numerator approaches 6 and denominator approaches 0, so the limit is $+\infty$.

Problem 2: Determine the value of $k$ that makes $f(x) = \begin{cases} \frac{x^2-4}{x-2} & \text{if } x \neq 2 \\ k & \text{if } x = 2 \end{cases}$ continuous at $x = 2$.

Solution: For continuity, we need $\lim_{x \to 2} f(x) = f(2) = k$.

Therefore, $k = 4$.

Problem 3: Evaluate $\lim_{x \to 0} \frac{\sin 5x}{\sin 3x}$.

Solution: Using the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$: