Limits, Continuity & Differentiability
Differentiability using definition
Grade 12
Question:
<p>Let \(f: R \to R\) be a function such that \(|f(x)| \leq x^2\), for all \(x \in R\). Then at \(x = 0\), \(f(x)\) is</p>
<p>continuous but not differentiable.</p>
<p>continuous as well as differentiable.</p>
<p>neither continuous nor differentiable.</p>
<p>differentiable but not continuous.</p>
Step-by-Step Solution
Key Concept: The squeeze theorem applied to absolute values: since |f(x)| ≤ x² and both -x² and x² approach 0 as x→0, f(x) must also approach 0. This forces continuity and differentiability at x=0.
<p><strong>Step 1:</strong> From the given condition |f(x)| ≤ x² for all x ∈ ℝ, substitute x = 0:</p><p>|f(0)| ≤ 0 ⟹ f(0) = 0</p><p><strong>Step 2:</strong> Check continuity at x = 0. Since |f(x)| ≤ x², we have:</p><p>-x² ≤ f(x) ≤ x²</p><p>By Squeeze Theorem: lim(x→0) f(x) = 0 = f(0), so f is continuous at x = 0.</p><p><strong>Step 3:</strong> Check differentiability at x = 0. Consider the difference quotient:</p><p>|f(x) - f(0)|/|x - 0| = |f(x)|/|x| ≤ x²/|x| = |x|</p><p>As x → 0, |x| → 0, so lim(x→0) f(x)/x = 0 by Squeeze Theorem.</p><p>Therefore f'(0) = 0 exists, making f differentiable at x = 0.</p><p><strong>Step 4:</strong> Since f is differentiable at x = 0, it is automatically continuous there.</p><p>∴ Answer: f(x) is continuous and differentiable at x = 0</p>
Correct Answer: B