Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>If \(f(x) = \begin{cases} e^{1/x} & x \neq 0 \\ 0 & x = 0 \end{cases}\) and \(4x + 1 = 28\), then which statement about \(f(x)\) is true?</p>
<p>(a) \(\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x)\) for \(x \neq 0\)</p>
<p>(b) <span style='font-style:italic;'>f</span>(<span style='font-style:italic;'>x</span>) is discontinuous at <span style='font-style:italic;'>x</span> = 3</p>
<p>(c) <span style='font-style:italic;'>f</span>(<span style='font-style:italic;'>x</span>) is discontinuous at <span style='font-style:italic;'>x</span> = 1</p>
<p>(d) <span style='font-style:italic;'>f</span>(<span style='font-style:italic;'>x</span>) is discontinuous at <span style='font-style:italic;'>x</span> = 1 and 3</p>
Step-by-Step Solution
Key Concept: Exponential functions are continuous at all points in their domain; discontinuities arise only at points of singularity.
<p>For $f(x) = e^{1/x}$ when $x \neq 0$, the function is continuous everywhere except possibly at <span style='font-style:italic;'>x</span> = 0.</p><p>At <span style='font-style:italic;'>x</span> = 0: $\lim_{x \to 0^+} e^{1/x} = +\infty$ and $\lim_{x \to 0^-} e^{1/x} = 0$</p><p>At <span style='font-style:italic;'>x</span> = 1 and <span style='font-style:italic;'>x</span> = 3, the function is continuous since these points are not singular.</p><p>∴ Answer is A.</p>
Correct Answer: A