Limits, Continuity & Differentiability
Differentiability at a point
Grade 12
Question:
<p>The function
\[f(x) = \begin{cases}
\frac{|x|(3e^{1/|x|} + 4)}{2 - e^{1/|x|}} & , x \neq 0 \\
0 & , x = 0
\end{cases}\]
is</p>
<p>(a) differentiable at \(x = 0\)</p>
<p>(b) non-differentiable at \(x = 0\)</p>
<p>(c) not continuous at \(x = 0\)</p>
<p>(d) cannot be determined</p>
Step-by-Step Solution
Key Concept: Check both continuity and differentiability by examining left and right derivatives separately using limit definition.
<p><strong>Solution:</strong> The given function may be written as</p>
$$f(x) = \begin{cases}
\frac{-x(3e^{-1/x} + 4)}{2 - e^{-1/x}} & , x < 0 \\
0 & , x = 0 \\
\frac{x(3e^{1/x} + 4)}{2 + e^{1/x}} & , x > 0
\end{cases}$$
<p>Analysis of the limit behavior shows that while the function may be continuous at $x = 0$, the derivative from the left and right do not match, making it non-differentiable at $x = 0$.</p>
<p>∴ Answer is (b) non-differentiable at $x = 0$</p>
Correct Answer: B