Limit Calculation

Grade 12 · calculus · 95 practice problems · read aloud

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Limit Calculation: The Foundation of Calculus 🧮

What is a Limit & Why is it Useful?

A limit describes the value that a function approaches as the input approaches some value. It's the fundamental concept behind calculus, allowing us to define instantaneous rates of change (derivatives) and areas under curves (integrals). Without limits, modern science and engineering wouldn't exist!

Step-by-Step Guide to Finding Limits

  1. Step 1: Direct Substitution - Plug the value (c) into the function. If you get a finite number, that's your limit!
  2. Step 2: Factor & Simplify - If you get an indeterminate form (like 0/0), try factoring and canceling common terms.
  3. Step 3: Rationalize - For limits with radicals, multiply by the conjugate to simplify.
  4. Step 4: Special Limits - Know key limits, like limx→0 (sin x)/x = 1.

Worked Examples

Example 1: Direct Substitution

Find limx→3 (2x² - 5)

Step 1: Substitute x = 3: 2(3)² - 5 = 2(9) - 5 = 18 - 5 = 13

Answer: 13 ✅

Example 2: Factoring (Indeterminate Form)

Find limx→2 (x² - 4)/(x - 2)

Step 1: Direct substitution gives 0/0. Indeterminate!

Step 2: Factor numerator: (x - 2)(x + 2)/(x - 2)

Step 3: Cancel (x - 2): limx→2 (x + 2)

Step 4: Substitute: 2 + 2 = 4

Answer: 4 ✅

🚨 Common Mistakes to Avoid

  • Canceling incorrectly: Only cancel common factors, not terms. (x+1)/(x+2) cannot be simplified to (1)/(2).
  • Assuming limit existence: A function doesn't always have a limit at a point. Check the left-hand and right-hand limits.
  • Misusing L'Hôpital's Rule: Only apply it to indeterminate forms like 0/0 or ∞/∞.

💡 Pro Tips & Tricks

  • Visualize it! Sketch a quick graph to see what y-value the function is approaching.
  • Know your conjugates: (a - b) and (a + b) are conjugates. Use them to simplify expressions with square roots.
  • Memorize the special limit: limx→0 sin(x)/x = 1. It's a lifesaver!

How to Practice

Start with limits by direct substitution, then move to factoring, then rationalization. Mix in problems that involve piecewise functions to test your understanding of one-sided limits. Try explaining the "why" behind each step to a friend—teaching is the best way to learn!

Practice problems

6 of the 95, worked through step by step — try them before opening the answer.

1 lim_{x→0} (sin(3x))/(2x) = ?

Hint: Consider the standard limit result for sine functions as the variable approaches zero, and how to manipulate the expression to match that form.

Show the answer

Answer: 1.5

  1. The problem is lim_{x→0} (sin(3x))/(2x).
  2. We know the standard limit: lim_{θ→0} (sin(θ))/(θ) = 1.
  3. To use this, we need the denominator to match the argument of sine. Multiply numerator and denominator by 3/3: (sin(3x))/(2x) = (3/2) × (sin(3x))/(3x).
  4. Now we have lim_{x→0} (3/2) × (sin(3x))/(3x).
  5. As x→0, 3x→0, so (sin(3x))/(3x) → 1.
  6. Therefore, the limit equals (3/2) × 1 = 3/2.
  7. 3/2 = 1.5

The answer is 1.5.

2 lim_(x→0) (sin(3x))/(2x) = ?

Hint: Consider the standard limit result for sine functions as the variable approaches zero.

Show the answer

Answer: 1.5

  1. We have the limit: lim_(x→0) (sin(3x))/(2x)
  2. We can rewrite this as: (1/2) × lim_(x→0) (sin(3x))/x
  3. Using the standard limit lim_(θ→0) sin(θ)/θ = 1, we need to make the denominator match the argument of sine
  4. Multiply numerator and denominator by 3: (1/2) × lim_(x→0) (3 × sin(3x))/(3x)
  5. This becomes: (1/2) × 3 × lim_(x→0) sin(3x)/(3x)
  6. As x→0, 3x→0, so lim_(x→0) sin(3x)/(3x) = 1
  7. Therefore, the limit equals (1/2) × 3 × 1 = 3/2 = 1.5

The answer is 1.5.

3 lim_(x→2) (x² - 4)/(x - 2) = ?

Hint: When evaluating limits that result in indeterminate forms, look for algebraic simplifications like factoring or rationalizing that can eliminate the problematic denominator.

Show the answer

Answer: 4

  1. Direct substitution If we substitute x = 2 into the expression, we get: (2² - 4)/(2 - 2) = (4 - 4)/(0) = 0/0 This is an indeterminate form, so we need to simplify the expression.
  2. Factor the numerator Notice that x² - 4 is a difference of squares: x² - 4 = (x - 2)(x + 2)
  3. Rewrite the expression So (x² - 4)/(x - 2) = [(x - 2)(x + 2)]/(x - 2) For x ≠ 2, we can cancel (x - 2) from numerator and denominator.
  4. Simplify After canceling, we get: x + 2
  5. Take the limit Now we take the limit as x approaches 2: lim_(x→2) (x + 2) = 2 + 2 = 4 Therefore, the limit is 4.

Let's solve the limit step-by-step. We are given: lim_(x→2) (x² - 4)/(x - 2)

4 lim_{x→2} (x² - 4)/(x - 2) = ?

Hint: When evaluating limits that result in indeterminate forms, look for algebraic simplifications like factoring that can eliminate the problematic denominator.

Show the answer

Answer: 4

  1. Direct substitution** If we substitute x = 2 into the expression, we get: (2² - 4)/(2 - 2) = (4 - 4)/(0) = 0/0 This is an indeterminate form, so we need to simplify the expression. --- **
  2. Factor the numerator** The numerator is x² - 4, which is a difference of squares: x² - 4 = (x - 2)(x + 2) So the expression becomes: (x² - 4)/(x - 2) = (x - 2)(x + 2)/(x - 2) --- **
  3. Cancel the common factor** For x ≠ 2, we can cancel (x - 2) from numerator and denominator: (x - 2)(x + 2)/(x - 2) = x + 2 So the original function is equivalent to x + 2 for all x except x = 2. --- **
  4. Take the limit** Now we take the limit as x approaches 2 of x + 2: lim_{x→2} (x + 2) = 2 + 2 = 4 --- **Final Answer:** 4

Let's solve the limit step-by-step. We are given: lim_{x→2} (x² - 4)/(x - 2) --- **

5 lim_{x→3} (x² - 9)/(x - 3) = ?

Hint: When evaluating limits that result in indeterminate forms, try algebraic manipulation to simplify the expression before substituting the limiting value.

Show the answer

Answer: 6

  1. Direct substitution** If we substitute x = 3 into the expression, we get: (3² - 9)/(3 - 3) = (9 - 9)/(0) = 0/0 This is an indeterminate form, so we need to simplify the expression. --- **
  2. Factor the numerator** Notice that x² - 9 is a difference of squares: x² - 9 = (x - 3)(x + 3) So the expression becomes: (x² - 9)/(x - 3) = (x - 3)(x + 3)/(x - 3) --- **
  3. Cancel the common factor** For x ≠ 3, we can cancel (x - 3) from numerator and denominator: (x - 3)(x + 3)/(x - 3) = x + 3 --- **
  4. Take the limit** Now we have: lim_{x→3} (x + 3) This is a simple polynomial, so we substitute x = 3: 3 + 3 = 6 --- **Final Answer:** 6

Let's solve the limit step by step. We are given: lim_{x→3} (x² - 9)/(x - 3) --- **

6 lim_{x→7} (x² - 49)/(x - 7) = ?

Hint: Notice that direct substitution gives 0/0. Factor the numerator as a difference of squares to cancel the common factor with the denominator.

Show the answer

Answer: 14

  1. Direct substitution gives (7² - 49)/(7 - 7) = (49 - 49)/0 = 0/0, an indeterminate form.
  2. Factor the numerator: x² - 49 = (x - 7)(x + 7).
  3. Rewrite the limit: lim_{x→7} ((x - 7)(x + 7))/(x - 7).
  4. Cancel the common factor (x - 7) (valid since x ≠ 7 in the limit): lim_{x→7} (x + 7).
  5. Evaluate by direct substitution: 7 + 7 = 14.

The answer is 14.

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