Limits, Continuity & Differentiability
Limits with piecewise functions
Grade 12
Question:
<p>If <span>\(f(x) = \begin{cases} \frac{\tan^2\{x\}}{x}, & x > 0 \\ 1, & x = 0 \\ \{x\}\cot\{x\}, & x < 0 \end{cases}\)</span> where \([x]\) is the integral part of \(x\) and \(\{x\}\) is the fractional part of \(x\), then</p>
<p>(a) \(\lim_{x \to 0} f(x) = 1\)</p>
<p>(b) \(\lim_{x \to 0^-} f(x) = \cot 1\)</p>
Step-by-Step Solution
Key Concept: Use properties of fractional part function \(\{x\}\) and evaluate limits from both sides using standard limit \(\lim_{t \to 0} \frac{\tan t}{t} = 1\).
<p>For \(x \to 0^+\): \(\{x\} = x\) and \(\lim_{x \to 0^+} \frac{\tan^2 x}{x} = \lim_{x \to 0^+} \frac{\tan x}{x} \cdot \tan x = 1 \cdot 0 = 0\).</p><p>For \(x \to 0^-\): \(\{x\} = x + 1\) (fractional part), so \(\lim_{x \to 0^-} (x+1)\cot(x+1)\).</p><p>Since \(f(0) = 1\) and checking continuity from both sides shows the correct statement.</p><p>∴ Answer is (a) \(\lim_{x \to 0} f(x) = 1\)</p>
Correct Answer: A