Limits, Continuity & Differentiability
Continuity and L'Hôpital's Rule
Grade 12
Question:
<p>The value of f at x = 0, so that function <span style='display:inline-block'>f(x) = \frac{2^x - 2^{-x}}{x}\)</span>, <span style='display:inline-block'>x \neq 0\)</span> is continuous at x = 0, is</p>
<p>(a) 0</p>
<p>(b) <span style='display:inline-block'>\log 4\)</span></p>
<p>(c) 4</p>
<p>(d) <span style='display:inline-block'>e^4\)</span></p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's Rule to evaluate the limit of an indeterminate form 0/0, then set this limit equal to f(0) for continuity.
<p><strong>Step 1:</strong> For continuity at x = 0, we need <span style='display:inline-block'>f(0) = \lim_{x \to 0} f(x)\)</span></p><p><strong>Step 2:</strong> <span style='display:inline-block'>\lim_{x \to 0} \frac{2^x - 2^{-x}}{x}\)</span> is of form 0/0</p><p><strong>Step 3:</strong> Apply L'Hôpital's Rule: <span style='display:inline-block'>\lim_{x \to 0} \frac{2^x \ln 2 - (-1)2^{-x}\ln 2}{1}\)</span></p><p><strong>Step 4:</strong> <span style='display:inline-block'>= \lim_{x \to 0} (2^x \ln 2 + 2^{-x}\ln 2) = \ln 2 + \ln 2 = 2\ln 2 = \ln 4 = \log 4\)</span></p><p>Therefore, <span style='display:inline-block'>f(0) = \log 4\)</span></p>
Correct Answer: B