Limits, Continuity & Differentiability
Limits of composite functions
Grade 12
Question:
<p>Let \(f(x) = \begin{cases} x+1; & x > 0 \\ 2-x; & x \leq 0 \end{cases}\) and \(g(x) = \begin{cases} 3+x; & x < 1 \\ x^2 - 2x - 2; & 1 \leq x < 2 \\ x-5; & x \geq 2 \end{cases}\) then: [Note: \([k]\) denotes greatest integer function less than or equal to \(k\).]</p>
<p>\(\lim_{x \to 0^+} g(f(x)) = -3\)</p>
<p>\(\lim_{x \to 0^-} g(f(x)) = -3\)</p>
<p>\(\lim_{x \to 0^+} [f(f(x))] = 0\)</p>
<p>\(\lim_{x \to 0^-} [g(g(x))] = -1\)</p>
Step-by-Step Solution
Key Concept: To check differentiability at x=0, verify that both left and right derivatives exist and are equal. The left derivative uses g(x)=2-x and right derivative uses f(x)=x+1.
<p><strong>Step 1: Check continuity at x=0</strong></p><p>f(0) = 2-0 = 2</p><p>lim(x→0⁻) f(x) = 2-0 = 2</p><p>lim(x→0⁺) f(x) = 0+1 = 1</p><p>Since lim(x→0⁻) ≠ lim(x→0⁺), f is <strong>not continuous</strong> at x=0.</p><p><strong>Step 2: Analyze the function more carefully</strong></p><p>The given piecewise function shows f(x) has a jump discontinuity at x=0 (jumps from 2 to 1).</p><p><strong>Step 3: Conclusion about differentiability</strong></p><p>Since f(x) is not continuous at x=0, it cannot be differentiable at x=0. A function must be continuous to be differentiable.</p><p>∴ Answer: A (The function is not differentiable at x=0)</p>
Correct Answer: A