Limits, Continuity & Differentiability
Continuity and L'Hospital's Rule
Grade 12
Question:
<p>Let \[f(x) = \frac{e^{\tan x} - e^x}{\tan x - x} + \frac{\log(\sec x + \tan x) - x}{\tan x - x}\] be a continuous function at \(x = 0\). The value of \(f(0)\) equals</p>
<p>(a) \(\frac{1}{2}\)</p>
<p>(b) \(\frac{2}{3}\)</p>
<p>(c) \(\frac{3}{2}\)</p>
<p>(d) \(2\)</p>
Step-by-Step Solution
Key Concept: Use continuity condition and L'Hospital's rule to evaluate the limit at x = 0.
<p><strong>Solution:</strong></p><p>For continuity at $x = 0$, we have</p><p>$$f(0) = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^{\tan x} - e^x}{\tan x - x} + \lim_{x \to 0} \frac{\log(\sec x + \tan x) - x}{\tan x - x}$$</p><p>Applying L'Hospital's rule:</p><p>$$= \lim_{x \to 0} \frac{e^{\tan x} \sec^2 x - e^x}{\sec^2 x - 1} + \lim_{x \to 0} \frac{\sec x - 1}{\sec^2 x - 1}$$</p><p>$$= 1 + \lim_{x \to 0} \frac{1}{\sec x + 1} = 1 + \frac{1}{2} = \frac{3}{2}$$</p><p>∴ Answer is (c) $\frac{3}{2}$</p>
Correct Answer: C