Limits, Continuity & Differentiability
1^∞, 0^0 and ∞^0 indeterminate forms
Grade 12
Question:
<p>If <span class="math-inline">\(\lim_{x \to 0} \left(1 + ax + bx^2\right)^{2/x} = e^3\)</span>, then</p>
<span class="math-inline">\(a = \dfrac{3}{2}\)</span> and <span class="math-inline">\(b \in \mathbb{R}\)</span>
<span class="math-inline">\(a = \dfrac{3}{2}\)</span> and <span class="math-inline">\(b \in \mathbb{R}^+\)</span>
<span class="math-inline">\(a = 0\)</span> and <span class="math-inline">\(b = 1\)</span>
<span class="math-inline">\(a = 1\)</span> and <span class="math-inline">\(b = 0\)</span>
Step-by-Step Solution
Key Concept: For limits of the form (1 + f(x))^(g(x)) where f(x) → 0, use the exponential form e^(lim g(x)·f(x)). Here, we need the exponent's numerator to match the logarithmic expansion of the base.
<p><strong>Step 1: Set up using exponential form</strong></p><p>For lim(x→0) [1 + ax + bx²]^(2/x) = e³, we use:</p><p>lim(x→0) [1 + u(x)]^(v(x)) = e^(lim(x→0) v(x)·u(x))</p><p>where u(x) = ax + bx² and v(x) = 2/x</p><p><strong>Step 2: Form the exponent</strong></p><p>The exponent becomes:</p><p>lim(x→0) (2/x)·(ax + bx²) = lim(x→0) (2ax/x + 2bx²/x) = lim(x→0) (2a + 2bx)</p><p><strong>Step 3: Evaluate the limit</strong></p><p>As x → 0: lim(x→0) (2a + 2bx) = 2a</p><p><strong>Step 4: Match with given result</strong></p><p>We need: e^(2a) = e³</p><p>Therefore: 2a = 3, which gives a = 3/2</p><p><strong>Step 5: Determine constraints on b</strong></p><p>The parameter b does not appear in the limiting exponent (the 2bx term vanishes as x → 0). Therefore, b can be any real number.</p><p><strong>Step 6: Verify answer</strong></p><p>The condition is: a = 3/2 and b ∈ ℝ, which corresponds to option A.</p><p><strong>∴ Answer:</strong> 1 (which is Option A)</p>
Correct Answer: 1