evaluate the limit
\\lim_{x \to 0} \frac{1 - \cos 3x}{x}\

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Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Trigonometric Limits", "LHopitals Rule" ], "new_concepts": [], "current_concepts": [ "Trigonometric Limits", "LHopitals Rule" ] } ## Check the limit form Using the **Trigonometric Limits** and **LHopitals Rule** knowledge points $$ \begin{aligned} &\lim_{x \to 0} (1 - \cos 3x) = 1 - \cos 0 = 0\ &\lim_{x \to 0} x = 0 \end{aligned} $$ ## Apply L'Hopital's Rule Using the **LHopitals Rule** knowledge point $$ \begin{aligned} &\frac{d}{dx}[1 - \cos 3x] = 3\sin 3x\ &\frac{d}{dx}[x] = 1\ &\lim_{x \to 0} \frac{1 - \cos 3x}{x} = \lim_{x \to 0} \frac{3\sin 3x}{1} = 3\sin 0 = 0 \end{aligned} $$ Evaluate the limit $$\lim_{x\to0} \frac{1 - \cos 3x}{x}$$ 0 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "LHopitals Rule" ] }

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Question

Explanation:

Check the limit form

Apply L'Hopital's Rule

Answer: