Introduction to Calculus: Detailed Notes & Exam Questions | Grade 12 Mathematics Unit 2
1. What is Calculus?
Dear student, welcome to Unit 2 of your Grade 12 Mathematics! In this unit, we begin our journey into Calculus — one of the most powerful and beautiful branches of mathematics.
Have you ever wondered how we can find the exact slope of a curved line? Or how we can calculate the area under a curve that isn’t a simple shape? These are exactly the kinds of problems calculus helps us solve.
Calculus is divided into two main parts:
- Differential Calculus — deals with rates of change and slopes of curves (we focus on this in Unit 2)
- Integral Calculus — deals with accumulation and areas under curves (covered later)
In this unit, we will start by understanding limits, which form the foundation of calculus. Then we will learn about continuity, and finally we will be introduced to derivatives.
Are you ready? Let’s begin step by step!
2. The Concept of a Limit
The idea of a limit is the starting point of all calculus. Let me explain it with a simple question:
What value does the function \( f(x) = x + 1 \) approach as \( x \) gets closer and closer to 3?
Well, if \( x = 2.9 \), then \( f(x) = 3.9 \). If \( x = 2.99 \), then \( f(x) = 3.99 \). If \( x = 2.999 \), then \( f(x) = 3.999 \). You can see that as \( x \) approaches 3, \( f(x) \) approaches 4.
We write this as:
We read this as: “the limit of x plus 1, as x approaches 3, equals 4.”
2.1 Understanding “Approaches” More Carefully
When we say \( x \to 3 \), we mean \( x \) gets closer and closer to 3, but \( x \) does not actually have to equal 3. In fact, the function might not even be defined at \( x = 3 \), but the limit can still exist!
Let me show you an important example. Consider:
At \( x = 3 \), this gives \( \frac{0}{0} \), which is undefined. So \( f(3) \) does not exist. But what about the limit?
Let’s simplify first:
Now as \( x \to 3 \), the expression \( x + 3 \) approaches 6. So:
Notice: The function was undefined at \( x = 3 \), but the limit still exists. This is a very important idea!
2.2 One-Sided Limits
Sometimes we need to approach a value from only one direction:
- Left-hand limit — \( x \) approaches the value from the left (from smaller values): \( \lim_{x \to a^-} \)
- Right-hand limit — \( x \) approaches the value from the right (from larger values): \( \lim_{x \to a^+} \)
The two-sided limit exists if and only if both one-sided limits exist and are equal:
Worked Example 1: Find \( \lim_{x \to 2^-} f(x) \) and \( \lim_{x \to 2^+} f(x) \) where:
Solution:
Left-hand limit: As \( x \to 2^- \), we use the first piece: \( 3(2) – 1 = 5 \)
Right-hand limit: As \( x \to 2^+ \), we use the second piece: \( 2 + 3 = 5 \)
Since both one-sided limits are equal to 5:
Think about it: What if the two one-sided limits were different? What would happen to the two-sided limit? Can you guess?
3. Properties and Theorems of Limits
To evaluate limits efficiently, we use the following properties. Assume \( \lim_{x \to a} f(x) = L \) and \( \lim_{x \to a} g(x) = M \):
| Property | Rule |
|---|---|
| Sum Rule | \(\lim_{x \to a} [f(x) + g(x)] = L + M\) |
| Difference Rule | \(\lim_{x \to a} [f(x) – g(x)] = L – M\) |
| Product Rule | \(\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M\) |
| Quotient Rule | \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}\), if \( M \neq 0 \) |
| Constant Rule | \(\lim_{x \to a} c = c\) |
| Power Rule | \(\lim_{x \to a} [f(x)]^n = L^n\) |
| Root Rule | \(\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}\), if \( L \geq 0 \) |
3.1 Direct Substitution
For many functions (polynomials, rational functions where denominator is not zero, root functions), we can simply substitute the value of \( a \) directly:
Worked Example 2: Evaluate \( \lim_{x \to -1} \frac{x^3 + 1}{x + 1} \)
Solution: Direct substitution gives \( \frac{0}{0} \) — indeterminate! So we factor:
Worked Example 3: Evaluate \( \lim_{x \to 0} \frac{\sqrt{x+4} – 2}{x} \)
Solution: Direct substitution gives \( \frac{0}{0} \). We rationalize the numerator by multiplying by \( \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} \):
4. Limits at Infinity
Sometimes we want to know what happens to a function as \( x \) becomes very large (goes to infinity) or very negative:
4.1 Polynomial Limits at Infinity
For a polynomial, the highest power of \( x \) dominates as \( x \to \pm\infty \):
The \( x^4 \) term grows much faster than the others.
4.2 Rational Function Limits at Infinity
For \( \frac{f(x)}{g(x)} \) where \( f \) and \( g \) are polynomials:
- If degree of numerator < degree of denominator: limit = 0
- If degree of numerator = degree of denominator: limit = ratio of leading coefficients
- If degree of numerator > degree of denominator: limit = \( \pm\infty \)
Worked Example 4: Find \( \lim_{x \to \infty} \frac{3x^2 + 5x – 1}{7x^2 – 2x + 4} \)
Solution: Both are degree 2, so divide numerator and denominator by \( x^2 \):
Worked Example 5: Find \( \lim_{x \to \infty} \frac{2x + 1}{x^2 – 3} \)
Solution: Numerator degree 1, denominator degree 2. Since 1 < 2:
5. Special Limits
There are two very important limits that you must memorize — they appear frequently in exams:
Worked Example 6: Evaluate \( \lim_{x \to 0} \frac{\sin 3x}{x} \)
Solution: We need to rewrite it to match the form \( \frac{\sin(\text{something})}{\text{something}} \). Multiply numerator and denominator by 3:
Worked Example 7: Evaluate \( \lim_{x \to 0} \frac{1 – \cos x}{x} \)
Solution: Multiply by \( \frac{1+\cos x}{1+\cos x} \):
Can you see how we used the special limit here? Practice makes perfect!
Evaluate: \( \lim_{x \to 0} \frac{\sin 5x}{\sin 2x} \)
Solution:
Evaluate: \( \lim_{x \to 0} \frac{\tan x}{x} \)
Solution: Since \( \tan x = \frac{\sin x}{\cos x} \):
6. Continuity
A function \( f \) is continuous at \( x = a \) if it satisfies three conditions:
- \( f(a) \) is defined (i.e., \( a \) is in the domain)
- \( \lim_{x \to a} f(x) \) exists
- \( \lim_{x \to a} f(x) = f(a) \)
If any one of these conditions fails, the function is discontinuous at \( x = a \).
Think of it this way: Can you draw the graph of the function through point \( x = a \) without lifting your pen? If yes, it’s continuous there!
6.1 Types of Discontinuity
| Type | Description | Example |
|---|---|---|
| Removable | Limit exists but \( f(a) \) is undefined or different from the limit | \( f(x)=\frac{x^2-1}{x-1} \) at \( x=1 \) |
| Jump | Left and right limits exist but are not equal | Piecewise functions with a gap |
| Infinite | Function goes to \( \pm\infty \) at the point | \( f(x)=\frac{1}{x} \) at \( x=0 \) |
Worked Example 8: Is \( f(x) = \frac{x^2 – 4}{x – 2} \) continuous at \( x = 2 \)?
Solution: Check the three conditions:
- \( f(2) = \frac{0}{0} \) — undefined. Condition 1 fails.
Since the first condition already fails, \( f \) is not continuous at \( x = 2 \). (This is a removable discontinuity, since the limit exists: \( \lim_{x \to 2} f(x) = 4 \).)
Worked Example 9: Is the following function continuous at \( x = 1 \)?
Solution:
- \( f(1) = 1^2 = 1 \) — defined ✓
- \( \lim_{x \to 1^-} f(x) = 1 \) and \( \lim_{x \to 1^+} f(x) = 2 \). Since \( 1 \neq 2 \), the limit does not exist. ✗
Condition 2 fails, so \( f \) is not continuous at \( x = 1 \). This is a jump discontinuity.
Find the value of \( k \) that makes \( f \) continuous at \( x = 3 \):
Solution: For continuity at \( x = 3 \), we need \( \lim_{x \to 3} f(x) = f(3) \).
Left-hand limit: \( \lim_{x \to 3^-} f(x) = k(3) + 1 = 3k + 1 \)
Right-hand limit: \( \lim_{x \to 3^+} f(x) = 3^2 – 5 = 4 \)
Set them equal: \( 3k + 1 = 4 \), so \( 3k = 3 \), giving \( k = 1 \).
7. The Derivative — Introduction
Now we come to the heart of this unit: the derivative. The derivative measures how fast a function is changing at any given point — it gives us the instantaneous rate of change.
Think about a car moving. Its speedometer tells you the speed at any instant — that’s like the derivative of position with respect to time!
7.1 Average Rate of Change
The average rate of change of \( f \) from \( x = a \) to \( x = b \) is:
This is exactly the slope of the secant line through points \( (a, f(a)) \) and \( (b, f(b)) \).
7.2 Instantaneous Rate of Change (The Derivative)
To find the rate of change at exactly one point, we let \( b \) get closer and closer to \( a \). This leads us to the definition of the derivative:
Here, \( h \) is a tiny change in \( x \). As \( h \to 0 \), the secant line becomes the tangent line, and its slope is the derivative!
7.3 Worked Examples — First Principles
Worked Example 10: Find \( f'(x) \) for \( f(x) = 3x^2 \) from first principles.
Solution:
So \( f'(x) = 6x \).
Worked Example 11: Find \( f'(x) \) for \( f(x) = \frac{1}{x} \) from first principles.
Solution:
Worked Example 12: Find \( f'(x) \) for \( f(x) = \sqrt{x} \) from first principles.
Solution:
Rationalize the numerator:
Use first principles to find \( f'(x) \) for \( f(x) = 2x^2 + 3x \).
Solution:
8. Differentiation Rules (Shortcuts)
While first principles always works, it can be slow. Mathematicians have developed shortcut rules:
8.1 Basic Rules
| Function | Derivative |
|---|---|
| \( c \) (constant) | \( 0 \) |
| \( x^n \) | \( nx^{n-1} \) (Power Rule) |
| \( c \cdot f(x) \) | \( c \cdot f'(x) \) (Constant Multiple) |
| \( f(x) \pm g(x) \) | \( f'(x) \pm g'(x) \) (Sum/Difference) |
Worked Example 13: Differentiate \( f(x) = 4x^5 – 3x^3 + 7x – 2 \)
Solution:
Worked Example 14: Differentiate \( f(x) = 3\sqrt{x} + \frac{5}{x^2} \)
Solution: First rewrite with fractional powers:
8.2 Product Rule
Worked Example 15: Differentiate \( f(x) = (2x+1)(x^2-3) \)
Solution: Let \( u = 2x+1 \), \( v = x^2-3 \). Then \( u’ = 2 \), \( v’ = 2x \).
8.3 Quotient Rule
Worked Example 16: Differentiate \( f(x) = \frac{x^2+1}{x-1} \)
Solution: Let \( u = x^2+1 \), \( v = x-1 \). Then \( u’ = 2x \), \( v’ = 1 \).
8.4 Chain Rule
Worked Example 17: Differentiate \( f(x) = (3x+1)^5 \)
Solution: Outer = \( (\cdot)^5 \), Inner = \( 3x+1 \).
Worked Example 18: Differentiate \( f(x) = \sqrt{2x^2 + 1} \)
Solution: Rewrite: \( f(x) = (2x^2+1)^{1/2} \)
- Sum/difference → Differentiate term by term
- Product of two functions → Product Rule
- Quotient of two functions → Quotient Rule
- Function inside a function → Chain Rule
- Sometimes you need a combination! e.g., quotient rule where the numerator or denominator needs the chain rule
9. Equation of Tangent and Normal Lines
One of the most important applications of derivatives is finding tangent and normal lines to curves.
9.1 Tangent Line
The tangent line to the curve \( y = f(x) \) at \( x = a \) has slope \( m = f'(a) \) and passes through \( (a, f(a)) \):
9.2 Normal Line
The normal line is perpendicular to the tangent line, so its slope is \( m_n = -\frac{1}{f'(a)} \) (provided \( f'(a) \neq 0 \)):
Worked Example 19: Find the equation of the tangent and normal to \( y = x^2 – 4x + 3 \) at \( x = 1 \).
Solution:
Step 1: Find \( y \)-coordinate: \( y = 1 – 4 + 3 = 0 \). Point is \( (1, 0) \).
Step 2: Find derivative: \( y’ = 2x – 4 \)
Step 3: Slope at \( x = 1 \): \( m = 2(1) – 4 = -2 \)
Step 4: Tangent line: \( y – 0 = -2(x – 1) \), so \( y = -2x + 2 \)
Step 5: Normal slope: \( m_n = \frac{1}{2} \)
Normal line: \( y – 0 = \frac{1}{2}(x – 1) \), so \( y = \frac{1}{2}x – \frac{1}{2} \) or \( 2y = x – 1 \)
Find the equation of the tangent line to \( y = \frac{1}{x} \) at \( x = 2 \).
Solution:
\( y \)-coordinate: \( y = \frac{1}{2} \). Point: \( (2, \frac{1}{2}) \)
Derivative: \( y’ = -\frac{1}{x^2} \). At \( x = 2 \): \( m = -\frac{1}{4} \)
Tangent: \( y – \frac{1}{2} = -\frac{1}{4}(x – 2) \)
10. Second Derivative
The second derivative is simply the derivative of the first derivative:
It tells us about the concavity of the function:
- If \( f”(x) > 0 \) on an interval → the graph is concave up (shaped like a cup ∪)
- If \( f”(x) < 0 \) on an interval → the graph is concave down (shaped like a cap ∩)
Worked Example 20: Find \( f”(x) \) for \( f(x) = 4x^3 – 6x^2 + 5 \).
Solution:
11. Summary of Key Exam Notes — Limits and Derivatives
- Limits: Always try direct substitution first. If you get \( \frac{0}{0} \), try factoring, rationalizing, or expanding.
- One-sided limits: If left limit ≠right limit, the two-sided limit does NOT exist.
- Limits at infinity: Compare degrees of numerator and denominator for rational functions.
- Special limits: Memorize \( \lim_{x \to 0}\frac{\sin x}{x} = 1 \) and \( \lim_{x \to \infty}(1+\frac{1}{x})^x = e \).
- Continuity: Three conditions — defined, limit exists, limit equals function value.
- First principles: Must show all steps for full marks. Don’t skip to the shortcut.
- Product Rule: “First × derivative of second + second × derivative of first”
- Quotient Rule: “Bottom × derivative of top − top × derivative of bottom, all over bottom squared”
- Chain Rule: Derivative of outer × derivative of inner
- Tangent/Normal: Find point, find derivative, get slope, write equation
(a) Evaluate \( \lim_{x \to 2} \frac{x^3 – 8}{x – 2} \)
(b) Hence, or otherwise, find \( \lim_{x \to 2} \frac{x^3 – 8}{x^2 – 4} \)
(a) Solution:
(b) Solution:
(Alternatively: \( \frac{x^3-8}{x^2-4} = \frac{x^3-8}{x-2} \cdot \frac{x-2}{x^2-4} \). The first factor → 12, second factor → \( \frac{0}{0} \to \frac{1}{4} \), so \( 12 \times \frac{1}{4} = 3 \).)
Use first principles to find the derivative of \( f(x) = x^3 \) and verify your answer using the power rule.
First Principles:
Power Rule verification: \( f(x) = x^3 \), so \( f'(x) = 3x^{3-1} = 3x^2 \). ✓ Both methods give the same answer!
Differentiate \( f(x) = \frac{(2x+1)^3}{\sqrt{x}} \) using the quotient rule and chain rule.
Solution: Let \( u = (2x+1)^3 \), \( v = \sqrt{x} = x^{1/2} \)
Multiply numerator and denominator by \( 2\sqrt{x} \):
Quick Revision Notes — Introduction to Calculus
1. Limit — Definition
2. Key Limit Properties
| Operation | Limit Rule |
|---|---|
| \( \lim [f \pm g]\) | \( \lim f \pm \lim g \) |
| \( \lim [f \cdot g]\) | \( \lim f \cdot \lim g \) |
| \( \lim \frac{f}{g}\) | \( \frac{\lim f}{\lim g} \) if \( \lim g \neq 0 \) |
| \( \lim [cf]\) | \( c \cdot \lim f \) |
| \( \lim [f^n]\) | \( (\lim f)^n \) |
3. Important Formulas to Memorize
4. Limits at Infinity — Quick Decision
5. Continuity — Three Conditions
- \( f(a) \) is defined
- \( \lim_{x \to a} f(x) \) exists
- \( \lim_{x \to a} f(x) = f(a) \)
6. Always-Continuous Functions
- Polynomials — continuous on \( (-\infty, \infty) \)
- Rational functions — continuous except where denominator = 0
- Root functions \( \sqrt[n]{f(x)} \) — continuous on their domain
- Trigonometric functions — continuous on their domains
- Exponential functions — continuous everywhere
7. Derivative — Definition (First Principles)
8. Differentiation Rules — Summary
| Rule | Formula |
|---|---|
| Constant | \( \frac{d}{dx}(c) = 0 \) |
| Power | \( \frac{d}{dx}(x^n) = nx^{n-1} \) |
| Constant Multiple | \( \frac{d}{dx}(cf) = c f’ \) |
| Sum/Difference | \( \frac{d}{dx}(f \pm g) = f’ \pm g’ \) |
| Product | \( \frac{d}{dx}(uv) = u’v + uv’ \) |
| Quotient | \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u’v – uv’}{v^2} \) |
| Chain | \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \) |
9. Common Derivatives
| \( f(x) \) | \( f'(x) \) |
|---|---|
| \( x^n \) | \( nx^{n-1} \) |
| \( \sqrt{x} \) | \( \frac{1}{2\sqrt{x}} \) |
| \( \frac{1}{x} \) | \( -\frac{1}{x^2} \) |
| \( \sin x \) | \( \cos x \) |
| \( \cos x \) | \( -\sin x \) |
| \( \tan x \) | \( \sec^2 x \) |
| \( e^x \) | \( e^x \) |
| \( \ln x \) | \( \frac{1}{x} \) |
10. Tangent and Normal Lines
11. Second Derivative & Concavity
- \( f”(x) > 0 \) → Concave up (∪ shape)
- \( f”(x) < 0 \) → Concave down (∩ shape)
- \( f”(x) = 0 \) → Possible point of inflection
12. Common Mistakes to Avoid
- Forgetting to check for \( \frac{0}{0} \) — Always try direct substitution first. If you get a number, you’re done!
- Wrong factoring — Double-check your factorizations, especially differences of cubes: \( a^3 – b^3 = (a-b)(a^2+ab+b^2) \)
- Mixing up product and quotient rules — Product: \( u’v + uv’ \). Quotient: \( \frac{u’v – uv’}{v^2} \). Note the MINUS in quotient!
- Forgetting the chain rule — When differentiating \( (2x+1)^5 \), you must multiply by the derivative of \( 2x+1 \), which is 2.
- First principles errors — Forgetting that \( h \to 0 \), or making algebra mistakes when expanding \( (x+h)^n \).
- Confusing limit with function value — \( \lim_{x \to a} f(x) \) is NOT the same as \( f(a) \). The limit may exist even when \( f(a) \) doesn’t.
- Normal slope sign error — Normal slope is \( -\frac{1}{m} \), not \( \frac{1}{m} \). Don’t forget the negative sign!
- Negative exponents — \( \frac{1}{x^2} = x^{-2} \), and its derivative is \( -2x^{-3} = \frac{-2}{x^3} \), not \( \frac{2}{x^3} \).
13. Quick Examples
Q: \( \lim_{x \to 5} (2x^2 – 3x + 1) \) = ?
A: Direct substitution: \( 2(25) – 15 + 1 = 36 \)
Q: \( \frac{d}{dx}(x^4 – 3x^2 + 2x) \) = ?
A: \( 4x^3 – 6x + 2 \)
Q: \( \frac{d}{dx}[\sin(3x)] \) = ?
A: \( \cos(3x) \cdot 3 = 3\cos(3x) \) (Chain rule!)
Q: \( \frac{d}{dx}\left(\frac{x}{x+1}\right) \) = ?
A: \( \frac{1\cdot(x+1) – x\cdot 1}{(x+1)^2} = \frac{1}{(x+1)^2} \)
Challenge Exam Questions — Introduction to Calculus
These questions are designed to test your deep understanding. Try each one before checking the answer!
Section A: Multiple Choice Questions
\( \lim_{x \to 0} \frac{\sin^2 x}{x^2} \) is equal to:
A) 0 B) 1 C) 2 D) Does not exist
Answer: B) 1
If \( f(x) = |x – 2| \), then \( \lim_{x \to 2} f(x) \) is:
A) 0 B) 2 C) -2 D) Does not exist
Answer: A) 0
Left-hand limit: \( \lim_{x \to 2^-} |x-2| = \lim_{x \to 2^-} (2-x) = 0 \)
Right-hand limit: \( \lim_{x \to 2^+} |x-2| = \lim_{x \to 2^+} (x-2) = 0 \)
Both equal 0, so \( \lim_{x \to 2} f(x) = 0 \).
Note: The limit exists even though \( f'(2) \) does not exist (the graph has a corner)!
\( \lim_{x \to \infty} \frac{5x^3 – 2x + 1}{3x^3 + x^2 – 7} \) is equal to:
A) 0 B) \( \frac{5}{3} \) C) \( \frac{3}{5} \) D) \( \infty \)
Answer: B) \( \frac{5}{3} \)
Both numerator and denominator are degree 3. The limit equals the ratio of leading coefficients: \( \frac{5}{3} \).
The derivative of \( f(x) = x^x \) (for \( x > 0 \)) is:
A) \( x \cdot x^{x-1} \) B) \( x^x \ln x \) C) \( x^x(1 + \ln x) \) D) \( x^x \cdot x \)
Answer: C) \( x^x(1 + \ln x) \)
Let \( y = x^x \). Take natural log: \( \ln y = x \ln x \).
Differentiate implicitly: \( \frac{1}{y} \cdot y’ = \ln x + x \cdot \frac{1}{x} = \ln x + 1 \)
This uses logarithmic differentiation — a powerful technique for variable exponents!
If \( f(x) = x^3 – 3x \), then the tangent line at \( x = 1 \) is:
A) \( y = 0 \) B) \( y = -2 \) C) \( y = x – 4 \) D) \( y = -2x \)
Answer: A) \( y = 0 \)
\( f(1) = 1 – 3 = -2 \). Point: \( (1, -2) \).
\( f'(x) = 3x^2 – 3 \). At \( x = 1 \): \( f'(1) = 3 – 3 = 0 \).
Equation: \( y – (-2) = 0(x – 1) \), so \( y + 2 = 0 \), giving \( y = -2 \).
Wait — that gives \( y = -2 \), which is option B! Let me recheck… Yes, the correct answer is B) \( y = -2 \).
Sorry for the confusion above. The slope is 0 (horizontal tangent), so the tangent line is the horizontal line \( y = -2 \).
Section B: Fill in the Blanks
\( \lim_{x \to 0} \frac{\sin 7x}{3x} \) = ________
Answer: \( \frac{7}{3} \)
If \( f(x) = (x^2 + 1)(3x – 2) \), then \( f'(0) \) = ________
Answer: -2
Using product rule: \( f'(x) = 2x(3x-2) + (x^2+1)(3) \)
Wait, let me recompute: \( f'(x) = 2x(3x-2) + (x^2+1)(3) \). At \( x=0 \): \( 0 + (0+1)(3) = 3 \).
The answer is 3. (I initially wrote -2 by mistake — always double-check!)
\( \lim_{x \to \infty} \frac{2x+3}{x^2+1} \) = ________
Answer: 0
Numerator degree (1) < Denominator degree (2), so the limit is 0.
The normal to the curve \( y = x^2 \) at \( x = 1 \) has slope ________
Answer: \( -\frac{1}{2} \)
\( y’ = 2x \). At \( x = 1 \): tangent slope = 2.
Normal slope = \( -\frac{1}{2} \).
Section C: Short Answer Questions
Find \( \lim_{x \to 4} \frac{\sqrt{x} – 2}{x – 4} \) without using L’Hôpital’s rule.
Solution:
Determine whether \( f(x) = \frac{x^2 – 1}{x^2 – 3x + 2} \) is continuous at \( x = 1 \).
Solution:
Factor: \( f(x) = \frac{(x-1)(x+1)}{(x-1)(x-2)} = \frac{x+1}{x-2} \) for \( x \neq 1 \).
At \( x = 1 \): Original function gives \( \frac{0}{0} \) — undefined. So condition 1 fails.
\( f \) is not continuous at \( x = 1 \). (It has a removable discontinuity since \( \lim_{x \to 1} f(x) = -2 \) exists.)
Differentiate \( f(x) = \sin^2(3x) \).
Solution: Use chain rule twice:
(Using the identity \( 2\sin A \cos A = \sin 2A \).)
Section D: Step-by-Step Calculation Questions
(a) Use first principles to find \( f'(x) \) for \( f(x) = \frac{1}{x+2} \).
(b) Hence find the equation of the tangent line at \( x = 0 \).
(a) First Principles:
(b) Tangent at \( x = 0 \):
\( f(0) = \frac{1}{2} \). Point: \( (0, \frac{1}{2}) \).
\( f'(0) = \frac{-1}{4} \)
Evaluate \( \lim_{x \to 0} \frac{\sin 3x – 3x}{x^3} \).
Solution: This requires knowing the series expansion of \( \sin \theta \):
So \( \sin 3x = 3x – \frac{(3x)^3}{6} + \cdots = 3x – \frac{27x^3}{6} + \cdots = 3x – \frac{9x^3}{2} + \cdots \)
This is a challenging question! If you haven’t learned series yet, you could also solve it using L’Hôpital’s rule (applying it three times), but the series method is more elegant.
Differentiate \( f(x) = \frac{x^2 + 1}{\sqrt{x^2 + 1}} \) and simplify your answer.
Solution: First, notice that \( f(x) = (x^2+1)^{1/2} \). This is much simpler!
Always simplify before differentiating — it saves a lot of work! If you had used the quotient rule directly, you’d get the same answer but with much more effort.
Find the value of \( k \) such that the function \( f \) is continuous everywhere:
Solution: The function is a polynomial on each piece, so it’s continuous on \( (-\infty, 2) \) and \( (2, \infty) \). We only need to check \( x = 2 \).
Condition: \( \lim_{x \to 2} f(x) = f(2) \)
\( f(2) = 4 + 2k + 3 = 2k + 7 \)
\( \lim_{x \to 2^+} f(x) = 4(2) + 1 = 9 \)
Set equal: \( 2k + 7 = 9 \), so \( 2k = 2 \), giving \( k = 1 \).
Find \( \lim_{x \to 0} \frac{e^x – 1 – x}{x^2} \).
Solution (using series expansion of \( e^x \)):
Alternative (L’Hôpital’s Rule): Direct substitution gives \( \frac{0}{0} \). Apply L’Hôpital once:
Still \( \frac{0}{0} \). Apply again:
The curve \( y = x^3 – 3x^2 + 2 \) has a tangent line that is parallel to the x-axis. Find the coordinates of the point(s) where this occurs and write the equation of the tangent line(s).
Solution: A tangent parallel to the x-axis has slope = 0.
At \( x = 0 \): \( y = 0 – 0 + 2 = 2 \). Point: \( (0, 2) \). Tangent: \( y = 2 \).
At \( x = 2 \): \( y = 8 – 12 + 2 = -2 \). Point: \( (2, -2) \). Tangent: \( y = -2 \).
Given \( f(x) = \frac{2x-1}{x+3} \), find:
(a) \( f'(x) \) using the quotient rule
(b) The value of \( x \) where the tangent line has slope \( \frac{1}{2} \)
(c) The equation of the normal line at that point
(a) Let \( u = 2x-1 \), \( v = x+3 \). Then \( u’ = 2 \), \( v’ = 1 \).
(b) Set \( f'(x) = \frac{1}{2} \):
(c) Using \( x = -3 + \sqrt{14} \):
Normal slope = \( -2 \) (negative reciprocal of \( \frac{1}{2} \))
(A similar normal line exists at the other point.)
Differentiate \( f(x) = \sqrt{\frac{x^2+1}{x^2-1}} \) using the chain rule and quotient rule together.
Solution: Let \( f(x) = \left(\frac{x^2+1}{x^2-1}\right)^{1/2} \)
By chain rule: \( f'(x) = \frac{1}{2}\left(\frac{x^2+1}{x^2-1}\right)^{-1/2} \cdot \frac{d}{dx}\left(\frac{x^2+1}{x^2-1}\right) \)
Now find the inner derivative by quotient rule:
Therefore:
This is a tough one — combining multiple rules is a key exam skill!
Prove that \( \lim_{x \to 0} \frac{\cos x – 1}{x^2} = -\frac{1}{2} \) without using L’Hôpital’s rule.
Solution: Multiply by \( \frac{\cos x + 1}{\cos x + 1} \):
If \( f(x) = ax^3 + bx^2 + cx + d \) and the tangent to the curve at \( x = 1 \) is horizontal, and the curve passes through \( (0, 5) \) and \( (1, 4) \), find \( a \), \( b \), and \( c \) in terms of \( d \). Then find the full equation if \( d = 5 \).
Solution:
Through \( (0, 5) \): \( d = 5 \)
Through \( (1, 4) \): \( a + b + c + d = 4 \), so \( a + b + c = -1 \) … (i)
Horizontal tangent at \( x = 1 \): \( f'(1) = 0 \)
Subtract (i) from (ii): \( 2a + b = 1 \), so \( b = 1 – 2a \)
From (i): \( c = -1 – a – b = -1 – a – (1-2a) = -2 + a \)
With \( d = 5 \): \( f(x) = ax^3 + (1-2a)x^2 + (a-2)x + 5 \)
We have one free parameter \( a \). To get a unique answer, we’d need one more condition. As stated, the family of curves is:
For example, if \( a = 1 \): \( f(x) = x^3 – x^2 – x + 5 \). You can verify: \( f(0)=5 \) ✓, \( f(1)=4 \) ✓, \( f'(1)=3-2-1=0 \) ✓.
Find \( \lim_{x \to 5} \frac{2x – 10}{\sqrt{x+4} – 3} \).
Solution: Direct substitution gives \( \frac{0}{0} \). Rationalize the denominator:
Find the second derivative of \( f(x) = \frac{x}{x^2+1} \) and evaluate \( f”(0) \).
Solution:
\( u = x \), \( v = x^2+1 \). \( u’ = 1 \), \( v’ = 2x \).
Now differentiate \( f'(x) = \frac{1-x^2}{(x^2+1)^2} \) using the quotient rule:
Let \( p = 1-x^2 \), \( q = (x^2+1)^2 \). Then \( p’ = -2x \), \( q’ = 2(x^2+1)(2x) = 4x(x^2+1) \).
Let \( f(x) = \begin{cases} x^2 \sin\frac{1}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \)
(a) Show that \( f \) is continuous at \( x = 0 \).
(b) Find \( f'(0) \) using the definition of the derivative.
(a) Continuity at \( x = 0 \):
Check: \( f(0) = 0 \) — defined ✓
We need \( \lim_{x \to 0} x^2 \sin\frac{1}{x} = 0 \).
Since \( -1 \leq \sin\frac{1}{x} \leq 1 \) for all \( x \neq 0 \), we have:
Both \( -x^2 \to 0 \) and \( x^2 \to 0 \) as \( x \to 0 \). By the Squeeze Theorem:
So \( f \) is continuous at \( x = 0 \). ✓
(b) Finding \( f'(0) \):
Again by Squeeze Theorem: \( -|h| \leq h \sin\frac{1}{h} \leq |h| \), and both bounds → 0.
This is a beautiful example! The function is continuous and differentiable at 0, even though \( \sin\frac{1}{x} \) oscillates wildly near 0. The \( x^2 \) factor “tames” the oscillation.