Limits, Continuity & Differentiability
Continuity and limit evaluation
Grade 12
<p>Given that \( f(x) = \dfrac{\sqrt{2}\cos x - 1}{\cot x - 1} \) is continuous at \( x = \dfrac{\pi}{4} \). Find the value of \( f\!\left(\dfrac{\pi}{4}\right) = k \).</p>
Step-by-Step Solution
Key Concept: Since f(x) is continuous at x = π/4, f(π/4) must equal lim(x→π/4) f(x). Use algebraic manipulation and L'Hôpital's rule or rationalization to evaluate this indeterminate 0/0 form.
<p><strong>Step 1:</strong> Check the form at x = π/4.<br/>cos(π/4) = 1/√2, so √2·cos(π/4) - 1 = √2·(1/√2) - 1 = 0<br/>cot(π/4) = 1, so cot(π/4) - 1 = 0<br/>This gives 0/0, an indeterminate form.</p><p><strong>Step 2:</strong> Apply L'Hôpital's rule:<br/>lim(x→π/4) [√2·cos(x) - 1]/[cot(x) - 1] = lim(x→π/4) [-√2·sin(x)]/[-csc²(x)]<br/>= lim(x→π/4) [√2·sin(x)·sin²(x)]/1<br/>= √2·sin(π/4)·sin²(π/4)<br/>= √2·(1/√2)·(1/√2)²<br/>= √2·(1/√2)·(1/2)<br/>= 1·(1/2) = 1/2</p><p><strong>Step 3:</strong> Since f is continuous at x = π/4, we have f(π/4) = lim(x→π/4) f(x) = 1/2</p><p>∴ k = <strong>1/2</strong></p>
Correct Answer: 1/2