Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>Given \(f(0) = 0\) and \(f(x) = \frac{1}{1-e^{-1/x}}\) for \(x \neq 0\). Then, only one of the following statements on \(f(x)\) is true.</p>
<p>(a) continuous at \(x = 0\)</p>
<p>(b) not continuous at \(x = 0\)</p>
<p>(c) both continuous and differentiable at \(x = 0\)</p>
<p>(d) not defined at \(x = 0\)</p>
Step-by-Step Solution
Key Concept: Check left and right limits separately; if they differ, the function is discontinuous.
<p>For continuity at $x = 0$, we need $\lim_{x \to 0} f(x) = f(0) = 0$.</p><p>As $x \to 0^+$: $e^{-1/x} \to 0$, so $f(x) = \frac{1}{1-0} = 1$</p><p>As $x \to 0^-$: $e^{-1/x} \to \infty$, so $f(x) = \frac{1}{1-\infty} = 0$</p><p>Since $\lim_{x \to 0^+} f(x) = 1 \neq 0 = \lim_{x \to 0^-} f(x)$, the limit does not exist.</p><p>∴ The function is not continuous at $x = 0$. Answer is B.</p>
Correct Answer: B