Limits, Continuity & Differentiability
Continuity at a Point
Grade 12
Question:
<p>Let <span class='formula'>f(x) = \begin{cases} \tan\left(\frac{\pi}{4} + x\right)^{1/x} & , x \neq 0 \\ k & , x = 0 \end{cases}</span></p><p>For what value of <span class='formula'>k</span> is <span class='formula'>f(x)</span> continuous at <span class='formula'>x = 0</span>?</p>
<p>(a) <span class='formula'>1</span></p>
<p>(b) <span class='formula'>e</span></p>
<p>(c) <span class='formula'>\frac{1}{e}</span></p>
<p>(d) <span class='formula'>e^2</span></p>
Step-by-Step Solution
Key Concept: For continuity at a point, the limit of the function as x approaches that point must equal the function value at that point. Use logarithmic differentiation and L'Hôpital's rule to evaluate the limit of an indeterminate form.
<p><strong>Solution:</strong></p><p>For continuity at <span class='formula'>x = 0</span>, we need <span class='formula'>\lim_{x \to 0} f(x) = f(0) = k</span></p><p>Calculate <span class='formula'>\lim_{x \to 0} \tan\left(\frac{\pi}{4} + x\right)^{1/x}</span></p><p><span class='formula'>= \lim_{x \to 0} \left(\frac{1 + \tan x}{1 - \tan x}\right)^{1/x}</span></p><p>Taking logarithm:</p><p><span class='formula'>\ln(L) = \lim_{x \to 0} \frac{1}{x} \ln\left(\frac{1 + \tan x}{1 - \tan x}\right)</span></p><p>Using L'Hôpital's rule and simplification:</p><p><span class='formula'>\ln(L) = \lim_{x \to 0} \frac{(1-\cos h) + \sin h}{2\sin h(\sin h + \cos h)} + \frac{h}{4\sin^2\frac{h}{2}\cos\frac{h}{2}}</span></p><p>After algebraic simplification:</p><p><span class='formula'>\ln(L) = \lim_{h \to 0} \frac{\sin\frac{h}{2} + \cos\frac{h}{2}}{2\cos\frac{h}{2}}(\sin h + \cos h) = 2</span></p><p>Therefore, <span class='formula'>L = e^2</span></p><p>∴ <span class='formula'>k = e^2</span></p>
Correct Answer: D