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:</p><p>[Note: \([k]\) denotes greatest integer function less than or equal to \(k\).]</p>
<p>\(\displaystyle\lim_{x\to 0^+} g(f(x)) = -3\)</p>
<p>\(\displaystyle\lim_{x\to 0^-} g(f(x)) = -3\)</p>
<p>\(\displaystyle\lim_{x\to 0^+} [f(f(x))] = 0\)</p>
<p>\(\displaystyle\lim_{x\to 0^-} [g(g(x))] = -1\)</p>
Step-by-Step Solution
Key Concept: Evaluate continuity and differentiability by checking left/right limits and derivatives at boundary points (x=0), and verify function values at specific points where the piecewise definition changes.
<p><strong>Step 1: Analyze f(x) at x=0</strong></p><p>For x>0: f(x)=x+1, so f(0⁺)=1</p><p>For x≤0: f(x)=2-x, so f(0)=2</p><p>Since f(0⁺)≠f(0), f is <strong>not continuous</strong> at x=0.</p><p><strong>Step 2: Check left and right derivatives of f at x=0</strong></p><p>Right derivative: f'(0⁺)=lim(h→0⁺)[f(h)-f(0)]/h = lim[(h+1-2)/h] = 1</p><p>Left derivative: f'(0⁻)=lim(h→0⁻)[f(h)-f(0)]/h = lim[(2-h-2)/h] = -1</p><p>Since f'(0⁺)≠f'(0⁻), <strong>both one-sided derivatives exist but are unequal</strong>.</p><p><strong>Step 3: Analyze g(x)</strong></p><p>For [x]<1 (i.e., x<1): g(x)=3+x</p><p>For [x]≥1 (i.e., x≥1): g(x)=4x</p><p>At x=1: g(1⁻)=3+1=4 and g(1)=4(1)=4, so g is continuous at x=1.</p><p><strong>Step 4: Verify statements:</strong></p><p><strong>A)</strong> Right derivative of f at x=0 is 1 ✓ TRUE</p><p><strong>B)</strong> f is continuous at x=0 ✗ FALSE (f(0⁺)=1≠f(0)=2)</p><p><strong>C)</strong> Left derivative of f at x=0 is -1 ✓ TRUE</p><p><strong>D)</strong> g is continuous at x=1 ✓ TRUE</p><p>∴ Answer: <strong>ACD</strong></p>
Correct Answer: ACD