Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>If <span style='font-style:italic;'>f</span>(<span style='font-style:italic;'>x</span>) = \begin{cases} \frac{1}{x} & x \neq 0 \\ 0 & x = 0 \end{cases}, then</p>
<p>(a) \(\lim_{x \to 0^+} f(x) = 1\)</p>
<p>(b) \(\lim_{x \to 0^-} f(x) = 0\)</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> = 0</p>
<p>(d) <span style='font-style:italic;'>f</span>(<span style='font-style:italic;'>x</span>) is continuous at <span style='font-style:italic;'>x</span> = 0</p>
Step-by-Step Solution
Key Concept: A function is continuous at a point if the left and right limits exist and equal the function value at that point.
<p>For continuity at <span style='font-style:italic;'>x</span> = 0, we need \(\lim_{x \to 0} f(x) = f(0)\).</p><p>\(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{x} = +\infty\) and \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{x} = -\infty\)</p><p>Since the limits do not exist and are not equal to <span style='font-style:italic;'>f</span>(0) = 0, the function is discontinuous at <span style='font-style:italic;'>x</span> = 0.</p><p>∴ Answer is C.</p>
Correct Answer: C