Limits & Continuity – Lessons with Examples

Limits are the gateway to all of calculus. Before derivatives, before integrals, before any of the “big” stuff — limits come first. And once you truly understand them, everything else in Calculus 1 starts to click.

This lesson breaks down limits from the ground up, with clear explanations, plenty of worked examples, and tips to handle every type you’ll see on your exam.

📚 Table of Contents

1. What is a Limit?
2. Limit Laws
3. Direct Substitution
4. Factoring Method
5. Rationalization Method
6. One-Sided Limits
7. Limits at Infinity
8. L’Hôpital’s Rule
9. Continuity
10. Common Mistakes
11. Practice Problems

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A limit asks a simple question: “What value does f(x) approach as x gets closer and closer to some number?”

The key idea: it doesn’t matter what happens exactly at that point — only what happens as you approach it.

The Formal Notation

$$\lim_{x \to a} f(x) = L$$

This reads: “The limit of f(x) as x approaches a equals L.”

A Simple Intuition

Imagine you’re walking toward a door. A limit asks: “Where are you heading?” — not whether you actually walk through the door.

Even if the door is locked (the function is undefined at that point), the limit still exists as long as you’re clearly heading toward the same spot from both sides.

✏️ Example 1 — Understanding a Limit Visually

Consider:

$$f(x) = \frac{x^2 – 1}{x – 1}$$

At x = 1, this is undefined (division by zero). But let’s see what happens as x approaches 1:

x	f(x)
0.9	1.9
0.99	1.99
0.999	1.999
1.001	2.001
1.01	2.01
1.1	2.1

Even though f(1) is undefined, f(x) is clearly approaching 2.

✅ Therefore:

$$\lim_{x \to 1} \frac{x^2 – 1}{x – 1} = 2$$

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These are the rules that let you break complicated limits into simpler pieces. You don’t need to memorize proofs — just know how to use them.

Law	Rule
Sum Rule	$$\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$
Difference Rule	$$\lim_{x \to a} [f(x) – g(x)] = \lim_{x \to a} f(x) – \lim_{x \to a} g(x)$$
Product Rule	$$\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$$
Quotient Rule	$$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}, \quad g(x) \neq 0$$
Constant Rule	$$\lim_{x \to a} c = c$$
Power Rule	$$\lim_{x \to a} [f(x)]^n = \left[\lim_{x \to a} f(x)\right]^n$$

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This is always your first move. Simply plug the value of x directly into the function. If you get a real number — you’re done!

When Does It Work?

• The function is a polynomial
• The function is a trig, exponential, or log function (and the point is in its domain)
• You do NOT get 0/0 or ∞/∞

✏️ Example 2

Problem:

$$\lim_{x \to 4} (3x^2 – 2x + 1)$$

Solution: Plug in x = 4:

$$3(4)^2 – 2(4) + 1 = 48 – 8 + 1 = 41$$

✅ Answer: 41

✏️ Example 3

Problem:

$$\lim_{x \to 0} \cos(x) + e^x$$

Solution: Plug in x = 0:

$$\cos(0) + e^0 = 1 + 1 = 2$$

✅ Answer: 2

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When direct substitution gives you 0/0 (called an indeterminate form), factoring is usually your best next move. Factor the numerator and denominator, cancel the common factor, then substitute.

✏️ Example 4

Problem:

$$\lim_{x \to 3} \frac{x^2 – 9}{x – 3}$$

Step 1: Try direct substitution:

$$\frac{9-9}{3-3} = \frac{0}{0}$$

❌ Indeterminate

Step 2: Factor the numerator:

$$\lim_{x \to 3} \frac{(x+3)(x-3)}{x-3}$$

Step 3: Cancel \( (x-3) \) and substitute:

$$\lim_{x \to 3} (x + 3) = 3 + 3 = 6$$

✅ Answer: 6

✏️ Example 5

Problem:

$$\lim_{x \to -2} \frac{x^2 + 5x + 6}{x + 2}$$

Step 1: Direct substitution → 0/0 ❌

Step 2: Factor:

$$\lim_{x \to -2} \frac{(x+2)(x+3)}{x+2}$$

Step 3: Cancel and substitute:

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

✅ Answer: 1

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When you see a square root in the numerator or denominator and get 0/0, multiply by the conjugate to eliminate the radical.

✏️ Example 6

Problem:

$$\lim_{x \to 0} \frac{\sqrt{x+4} – 2}{x}$$

Step 1: Direct substitution → 0/0 ❌

Step 2: Multiply by the conjugate:

$$\lim_{x \to 0} \frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)}$$

Step 3: Simplify the numerator using difference of squares:

$$\lim_{x \to 0} \frac{(x+4) – 4}{x(\sqrt{x+4}+2)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+4}+2)}$$

Step 4: Cancel x and substitute:

$$\lim_{x \to 0} \frac{1}{\sqrt{x+4}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{4}$$

✅ Answer: 1/4

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Sometimes a function approaches different values depending on which direction you come from. That’s where one-sided limits come in.

• $$\lim_{x \to a^-} f(x)$$ → approaching from the left
• $$\lim_{x \to a^+} f(x)$$ → approaching from the right

The Golden Rule

$$\lim_{x \to a} f(x) \text{ exists} \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$$

If the left and right limits are not equal, the limit Does Not Exist (DNE).

✏️ Example 7

Problem: Given

$$f(x) = \begin{cases} x + 1 & x < 2 \\ 3x - 2 & x \geq 2 \end{cases}$$

Find \( \lim_{x \to 2} f(x) \)

Left limit:

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

Right limit:

$$\lim_{x \to 2^+} (3x-2) = 4$$

Since 3 ≠ 4:

✅ Answer: DNE (Does Not Exist)

✏️ Example 8

Problem: Given

$$f(x) = \begin{cases} 2x + 1 & x < 3 \\ x + 4 & x \geq 3 \end{cases}$$

Find \( \lim_{x \to 3} f(x) \)

Left limit:

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

Right limit:

$$\lim_{x \to 3^+} (x+4) = 7$$

Since both equal 7:

✅ Answer: 7

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Limits at infinity ask: “What does f(x) approach as x grows without bound?” These show up constantly on exams.

The Key Strategy

Divide every term by the highest power of x in the denominator, then use the fact that:

$$\lim_{x \to \infty} \frac{1}{x^n} = 0 \quad \text{for any } n > 0$$

Quick Rules for Rational Functions

• Degree of numerator < denominator → Limit = 0
• Degree of numerator = denominator → Limit = ratio of leading coefficients
• Degree of numerator > denominator → Limit = ±∞ (DNE)

✏️ Example 9

Problem:

$$\lim_{x \to \infty} \frac{5x^2 + 3x}{2x^2 – 7}$$

Same degree → divide by \( x^2 \):

$$\lim_{x \to \infty} \frac{5 + \frac{3}{x}}{2 – \frac{7}{x^2}} = \frac{5 + 0}{2 – 0} = \frac{5}{2}$$

✅ Answer: 5/2

✏️ Example 10

Problem:

$$\lim_{x \to \infty} \frac{4x + 1}{x^2 + 3}$$

Numerator degree (1) < denominator degree (2):

✅ Answer: 0

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When you get \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) and factoring doesn’t work, L’Hôpital’s Rule is your best friend.

The Rule

If \( \lim_{x \to a} \frac{f(x)}{g(x)} \) gives \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

⚠️ Important: You differentiate the top and bottom separately — this is NOT the quotient rule!

✏️ Example 11

Problem:

$$\lim_{x \to 0} \frac{\sin(x)}{x}$$

Direct substitution → 0/0 ❌

Apply L’Hôpital’s Rule:

$$\lim_{x \to 0} \frac{\cos(x)}{1} = \cos(0) = 1$$

✅ Answer: 1

✏️ Example 12

Problem:

$$\lim_{x \to \infty} \frac{x^2}{e^x}$$

Gives ∞/∞ → Apply L’Hôpital’s Rule:

$$\lim_{x \to \infty} \frac{2x}{e^x}$$

Still ∞/∞ → Apply again:

$$\lim_{x \to \infty} \frac{2}{e^x} = 0$$

✅ Answer: 0

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A function is continuous at x = a if there’s no break, jump, or hole at that point. Formally, three conditions must all be true:

1. \( f(a) \) is defined
2. \( \lim_{x \to a} f(x) \) exists
3. \( \lim_{x \to a} f(x) = f(a) \)

If any one of these fails, the function is discontinuous at that point.

Types of Discontinuity

Type	What It Looks Like	Example
Removable (Hole)	A single point missing from the graph	\( \frac{x^2-1}{x-1} \) at x = 1
Jump	Left and right limits exist but are different	Piecewise functions
Infinite	Function goes to ±∞	\( \frac{1}{x} \) at x = 0

✏️ Example 13

Problem: Is \( f(x) = \frac{x^2 – 4}{x – 2} \) continuous at x = 2?

Check condition 1: f(2) = 0/0 → undefined ❌

Since condition 1 fails, f(x) is not continuous at x = 2.

It has a removable discontinuity (hole) at x = 2.

✅ Answer: Not continuous — removable discontinuity at x = 2

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• ❌ Canceling before checking: Always try direct substitution first
• ❌ Using L’Hôpital’s Rule when it doesn’t apply: Only use it for 0/0 or ∞/∞
• ❌ Confusing the limit with the function value: The limit is what f(x) approaches, not necessarily f(a)
• ❌ Forgetting one-sided limits: Always check both sides for piecewise functions
• ❌ Applying quotient rule in L’Hôpital’s: Differentiate top and bottom separately!
• ❌ Saying limit = ∞ “exists”: If a limit equals ±∞, it technically Does Not Exist

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Try these on your own before checking the answers!

Evaluate each limit:

1.

$$\lim_{x \to 5} (2x^2 – 3x + 1)$$

2.

$$\lim_{x \to 4} \frac{x^2 – 16}{x – 4}$$

3.

$$\lim_{x \to 0} \frac{\sqrt{x + 9} – 3}{x}$$

4.

$$\lim_{x \to \infty} \frac{6x^3 – 2x}{3x^3 + x^2}$$

5.

$$\lim_{x \to 0} \frac{\tan(x)}{x}$$

Answers:

1. 36  |  2. 8  |  3. 1/6  |  4. 2  |  5. 1

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Good luck — limits are the foundation, and you’ve got them covered! 🎓

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