Limits, Continuity & Differentiability
General
Grade 12
Question:
<div class="math-display">\[ \lim_{x \to 2} \frac{\sqrt{1 - \cos{2(x - 2)}}}{x - 2} \]</div>
equals -<span class="math-inline">\( \sqrt{2} \)</span>
equals <span class="math-inline">\( \frac{1}{\sqrt{2}} \)</span>
does not exist
equals <span class="math-inline">\( \sqrt{2} \)</span>
Step-by-Step Solution
Key Concept: Use the trigonometric identity √(1 - cos(2θ)) = √2|sin(θ)| to simplify the numerator, then evaluate the limit by recognizing the left and right limits separately since the absolute value creates a piecewise behavior.
<p><strong>Step 1:</strong> Let u = x - 2, so as x → 2, we have u → 0. The limit becomes:<br/>$$\lim_{u \to 0} \frac{\sqrt{1 - \cos(2u)}}{u}$$</p><p><strong>Step 2:</strong> Apply the trigonometric identity 1 - cos(2u) = 2sin²(u):<br/>$$\sqrt{1 - \cos(2u)} = \sqrt{2\sin^2(u)} = \sqrt{2}|\sin(u)|$$</p><p><strong>Step 3:</strong> Substitute into the limit:<br/>$$\lim_{u \to 0} \frac{\sqrt{2}|\sin(u)|}{u}$$</p><p><strong>Step 4:</strong> Evaluate the right-hand limit (u → 0⁺):<br/>For u > 0: |sin(u)| = sin(u), so<br/>$$\lim_{u \to 0^+} \frac{\sqrt{2}\sin(u)}{u} = \sqrt{2} \cdot 1 = \sqrt{2}$$</p><p><strong>Step 5:</strong> Evaluate the left-hand limit (u → 0⁻):<br/>For u < 0: |sin(u)| = -sin(u), so<br/>$$\lim_{u \to 0^-} \frac{\sqrt{2}(-\sin(u))}{u} = \lim_{u \to 0^-} \frac{-\sqrt{2}\sin(u)}{u} = -\sqrt{2}$$</p><p><strong>Step 6:</strong> Since the left-hand limit (-√2) ≠ right-hand limit (√2), the limit does not exist.<br/>$$\therefore \text{Answer: 4 (C: does not exist)}$$</p>
Correct Answer: 4