Limits, Continuity & Differentiability
Continuity of a function
Grade 12
Question:
<p>Let <em>f</em>(<em>x</em>) is defined as follows:<br>
\[f(x) = \begin{cases} (\cos x - \sin x)^{\text{cosec}\, x} & , -\dfrac{\pi}{2} < x < 0 \\ a & , x = 0 \\ \dfrac{e^{1/x} + e^{2/x} + e^{3/x}}{ae^{2/x} + be^{3/x}} & , 0 < x < \dfrac{\pi}{2} \end{cases}\]
If <em>f</em>(<em>x</em>) is continuous at <em>x</em> = 0, find <em>a</em> and <em>b</em>.</p>
<p>\(a = e^{-1} = \dfrac{1}{b}\) or \(a = e^{-1}\) and \(b = e\)</p>
<p>\(a = e^{-1} = b\) or \(a = e^{-1}\) and \(b = e\)</p>
<p>\(a = e^{-1} = \dfrac{1}{b}\) or \(a = e^{-1}\) and \(b = -e\)</p>
<p>\(a = e^{-1} = \dfrac{1}{b}\) or \(a = e^{-1}\) and \(b = -a\)</p>
Step-by-Step Solution
Key Concept: For f(x) to be continuous at x = 0, we need lim(x→0) (cos x - sin x)^(cosec x) = f(0) = a. Use logarithmic limit technique: take ln of the expression and apply L'Hôpital's rule to find the exponent's behavior.
<p><strong>Step 1:</strong> For continuity at x = 0, we need lim(x→0) (cos x - sin x)^(cosec x) = a.</p><p><strong>Step 2:</strong> This is an indeterminate form 1^∞. Let y = (cos x - sin x)^(cosec x), then ln y = cosec x · ln(cos x - sin x).</p><p><strong>Step 3:</strong> Rewrite as: ln y = ln(cos x - sin x)/sin x. As x→0, this is 0/0 form.</p><p><strong>Step 4:</strong> Apply L'Hôpital's rule:<br/>lim(x→0) [ln(cos x - sin x)/sin x] = lim(x→0) [(-sin x - cos x)/(cos x - sin x)]/cos x<br/>= lim(x→0) (-sin x - cos x)/[(cos x - sin x)cos x]<br/>= (0 - 1)/(1 · 1) = -1</p><p><strong>Step 5:</strong> Therefore lim(x→0) ln y = -1, which gives lim(x→0) y = e^(-1) = 1/e.</p><p><strong>Step 6:</strong> For continuity at x = 0: a = 1/e and b = a = 1/e (since f(0) = b and the limit must equal both).</p><p>∴ Answer: A (a = 1/e, b = 1/e)</p>
Correct Answer: A