Limits, Continuity & Differentiability
Discontinuity at a Point
Grade 12
Question:
<p>If <span class="math">f(x) = \begin{cases} 1+e^{1/x} & , x < 0 \\ 0 & , x = 0 \end{cases}</span>, then</p>
<p>(a) <span class="math">\lim_{x \to 0^+} f(x) = 0</span></p>
<p>(b) <span class="math">\lim_{x \to 0^-} f(x) = 1</span></p>
<p>(c) <span class="math">f(x)</span> is discontinuous at <span class="math">x=0</span></p>
<p>(d) <span class="math">f(x)</span> is continuous at <span class="math">x=0</span></p>
Step-by-Step Solution
Key Concept: Check left and right limits and compare with the function value at the point to determine continuity.
<p><strong>Solution:</strong> For <span class="math">x < 0</span>, <span class="math">f(x) = 1 + e^{1/x}</span>. As <span class="math">x \to 0^-</span>, <span class="math">\frac{1}{x} \to -\infty</span>, so <span class="math">e^{1/x} \to 0</span>, thus <span class="math">\lim_{x \to 0^-} f(x) = 1</span>. The function is not defined for <span class="math">x > 0</span>, and <span class="math">f(0) = 0</span>. Since <span class="math">\lim_{x \to 0^-} f(x) = 1 \neq 0 = f(0)</span>, the function is discontinuous at <span class="math">x=0</span>.</p>
Correct Answer: c