By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Indeterminate forms are expressions that take the form of 0/0, ∞/∞, 0×∞, ∞-∞, 0^0, 1^∞, or ∞^0, which cannot be evaluated directly. This topic appears in exams to test your ability to resolve these forms using techniques like factoring, rationalizing, and trigonometric identities. Typical questions involve limits and require you to manipulate the expression into a determinate form.
This topic is frequently tested in calculus exams, particularly in sections on limits and continuity. It appears in about 10-20% of questions and can carry 5-10 marks each. It tests your analytical skills and understanding of algebraic manipulation and trigonometric identities.
If you are missing these, you will struggle with factoring complex polynomials, applying trig identities correctly, and understanding the concept of limits.
To resolve an indeterminate form 0/0, you need to manipulate the expression into a form where the limit can be directly evaluated.
Think of the indeterminate form as a locked door. Factoring, rationalizing, and trig identities are your keys to unlock it.
Intermediate
Question: Evaluate the limit: [ \lim_{x \to 0} \frac{x^2 - x}{x} ]
Step-by-Step: 1. Factor the numerator: ( x^2 - x = x(x - 1) ).2. Simplify: ( \frac{x(x - 1)}{x} = x - 1 ).3. Evaluate the limit: ( \lim_{x \to 0} (x - 1) = -1 ).
Answer: -1
Question: Evaluate the limit: [ \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} ]
Step-by-Step: 1. Rationalize the numerator: Multiply by the conjugate ( \sqrt{x+1} + 1 ).2. Simplify: ( \frac{(\sqrt{x+1} - 1)(\sqrt{x+1} + 1)}{x(\sqrt{x+1} + 1)} = \frac{x}{x(\sqrt{x+1} + 1)} = \frac{1}{\sqrt{x+1} + 1} ).3. Evaluate the limit: ( \lim_{x \to 0} \frac{1}{\sqrt{x+1} + 1} = \frac{1}{2} ).
Answer: 1/2
Question: Evaluate the limit: [ \lim_{x \to 0} \frac{\sin(x)}{x} ]
Step-by-Step: 1. Recognize the trigonometric identity: ( \sin(x) \approx x ) as ( x \to 0 ).2. Simplify: ( \frac{\sin(x)}{x} \approx \frac{x}{x} = 1 ).3. Evaluate the limit: ( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 ).
Answer: 1
Correct Approach: Ensure all common terms are factored out.
Mistake: Forgetting to rationalize.
Correct Approach: Multiply by the conjugate to eliminate irrationalities.
Mistake: Misapplying trigonometric identities.
Correct Approach: Use the correct identity for the given expression.
Mistake: Overlooking L'Hospital's Rule.
Favored By: Calculus exams
Proof Questions: Requires proving a statement involving indeterminate forms.
Favored By: Advanced calculus exams
Multiple-Choice: Options include correct and incorrect simplifications.
Question: Evaluate the limit: [ \lim_{x \to 0} \frac{x^3 - x^2}{x^2} ]
Options: A) 0 B) -1 C) 1 D) 2
Correct Answer: B) -1
Explanation: Factor the numerator: ( x^3 - x^2 = x^2(x - 1) ). Simplify: ( \frac{x^2(x - 1)}{x^2} = x - 1 ). Evaluate the limit: ( \lim_{x \to 0} (x - 1) = -1 ).
Why the Distractors Are Tempting: - A) 0: Looks plausible if you misapply the limit.- C) 1: Might seem correct if you incorrectly simplify.- D) 2: Could be chosen if you make a calculation error.
Question: Evaluate the limit: [ \lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} ]
Options: A) 1/4 B) 1/2 C) 1 D) 2
Correct Answer: A) 1/4
Explanation: Rationalize the numerator: Multiply by the conjugate ( \sqrt{x+4} + 2 ). Simplify: ( \frac{(\sqrt{x+4} - 2)(\sqrt{x+4} + 2)}{x(\sqrt{x+4} + 2)} = \frac{x}{x(\sqrt{x+4} + 2)} = \frac{1}{\sqrt{x+4} + 2} ). Evaluate the limit: ( \lim_{x \to 0} \frac{1}{\sqrt{x+4} + 2} = \frac{1}{4} ).
Why the Distractors Are Tempting: - B) 1/2: Might seem correct if you misapply the conjugate.- C) 1: Could be chosen if you simplify incorrectly.- D) 2: Looks plausible if you make a calculation error.
Question: Evaluate the limit: [ \lim_{x \to 0} \frac{\tan(x)}{x} ]
Options: A) 0 B) 1/2 C) 1 D) 2
Correct Answer: C) 1
Explanation: Recognize the trigonometric identity: ( \tan(x) \approx x ) as ( x \to 0 ). Simplify: ( \frac{\tan(x)}{x} \approx \frac{x}{x} = 1 ). Evaluate the limit: ( \lim_{x \to 0} \frac{\tan(x)}{x} = 1 ).
Why the Distractors Are Tempting: - A) 0: Looks plausible if you misapply the limit.- B) 1/2: Might seem correct if you misapply the identity.- D) 2: Could be chosen if you make a calculation error.
Question: Evaluate the limit: [ \lim_{x \to 0} \frac{x^2}{x - x^2} ]
Correct Answer: A) 0
Explanation: Factor the denominator: ( x - x^2 = x(1 - x) ). Simplify: ( \frac{x^2}{x(1 - x)} = \frac{x}{1 - x} ). Evaluate the limit: ( \lim_{x \to 0} \frac{x}{1 - x} = 0 ).
Why the Distractors Are Tempting: - B) -1: Might seem correct if you misapply the limit.- C) 1: Could be chosen if you simplify incorrectly.- D) 2: Looks plausible if you make a calculation error.
Question: Evaluate the limit: [ \lim_{x \to 0} \frac{\sin(2x)}{x} ]
Options: A) 0 B) 1 C) 2 D) 4
Correct Answer: C) 2
Explanation: Recognize the trigonometric identity: ( \sin(2x) \approx 2x ) as ( x \to 0 ). Simplify: ( \frac{\sin(2x)}{x} \approx \frac{2x}{x} = 2 ). Evaluate the limit: ( \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 ).
Why the Distractors Are Tempting: - A) 0: Looks plausible if you misapply the limit.- B) 1: Might seem correct if you misapply the identity.- D) 4: Could be chosen if you make a calculation error.
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