Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>If \(f(x) = \begin{cases} \frac{9e^{1/x} - e^{-1/x}}{x^2 e^{1/x}} & x \neq 0 \\ 0 & x = 0 \end{cases}\), then at \(x = 0\), \(f(x)\) is</p>
<p>(a) differentiable</p>
<p>(b) not differentiable</p>
<p>(c) \(f'(0) = -1\)</p>
<p>(d) \(f'(0) = 1\)</p>
Step-by-Step Solution
Key Concept: Exponential functions with $1/x$ in the exponent cause the function to behave wildly near the origin, preventing differentiability even if we define $f(0)=0$.
<p><strong>Simplification:</strong> $f(x) = \frac{9e^{1/x} - e^{-1/x}}{x^2 e^{1/x}} = \frac{9}{x^2} - \frac{e^{-2/x}}{x^2}$</p><p><strong>Behavior near $x = 0$:</strong> As $x \to 0$, $\frac{9}{x^2} \to \infty$ and $\frac{e^{-2/x}}{x^2}$ oscillates and diverges. The function does not have a well-defined limit as $x \to 0$.</p><p><strong>For differentiability:</strong> $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h}$, which involves $\frac{1}{h^3}$ terms that diverge. The derivative does not exist.</p><p>∴ Answer is (b).</p>
Correct Answer: b