y = f(x) is a continuous function such that its graph passes through (a,0). Then Lim <span class="math-inline">\( x \to a \)</span> <span class="math-inline">\( \frac{\log_e(1+3f(x))}{2f(x)} \)</span> is-
1
0
<span class="math-inline">\( \frac{3}{2} \)</span>
<span class="math-inline">\( \frac{2}{3} \)</span>
Step-by-Step Solution
Key Concept: Since f(x) is continuous and passes through (a,0), we have f(a) = 0. This makes the limit indeterminate (0/0 form), requiring us to use the standard logarithmic limit: lim(u→0) ln(1+u)/u = 1.
<p><strong>Step 1:</strong> Identify the given information. f(x) is continuous and its graph passes through (a,0), which means f(a) = 0.</p><p><strong>Step 2:</strong> Evaluate the limit by direct substitution to check the form:</p><p>At x = a: numerator = ln(1+3f(a)) = ln(1+0) = 0</p><p>At x = a: denominator = 2f(a) = 0</p><p>This is a 0/0 indeterminate form.</p><p><strong>Step 3:</strong> Rewrite the limit algebraically by introducing a standard form:</p><p>$$\lim_{x \to a} \frac{\ln(1+3f(x))}{2f(x)} = \lim_{x \to a} \frac{\ln(1+3f(x))}{3f(x)} \cdot \frac{3f(x)}{2f(x)}$$</p><p><strong>Step 4:</strong> Simplify the second fraction:</p><p>$$= \lim_{x \to a} \frac{\ln(1+3f(x))}{3f(x)} \cdot \frac{3}{2}$$</p><p><strong>Step 5:</strong> As x → a, f(x) → f(a) = 0 (by continuity), so let u = 3f(x) → 0.</p><p><strong>Step 6:</strong> Apply the standard limit formula lim(u→0) ln(1+u)/u = 1:</p><p>$$= 1 \cdot \frac{3}{2} = \frac{3}{2}$$</p><p><strong>∴ Answer:</strong> y = {x, 0 ≤ x ≤ 1; 1 < x ≤ 2}</p>
Correct Answer: y = {x, 0 ≤ x ≤ 1; 1 < x ≤ 2}