<p>State whether the statement is true or false: \(\lim_{y \to 0} \frac{\log(1+x)}{3^y - 1} = \frac{1}{\log_e 3}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the limit involves y→0 in the denominator while the numerator contains only x (a constant). The numerator log(1+x) is independent of y, so only the denominator's behavior as y→0 determines the limit's existence.
<p><strong>Step 1:</strong> Analyze the structure of the limit as y→0.</p><p>The numerator log(1+x) is independent of y and is a constant (assuming x is fixed and x > -1).</p><p><strong>Step 2:</strong> Evaluate the denominator as y→0.</p><p>We know that lim(y→0) 3^y = 1, so lim(y→0) (3^y - 1) = 0.</p><p><strong>Step 3:</strong> Determine the limit form.</p><p>$$\lim_{y \to 0} \frac{\log(1+x)}{3^y - 1} = \frac{\log(1+x)}{0}$$</p><p>Since the numerator is a non-zero constant and the denominator approaches 0, the limit is <strong>undefined (±∞)</strong>, not equal to 1/log_e 3.</p><p><strong>Step 4:</strong> Conclusion.</p><p>The given statement is <strong>FALSE</strong>. The limit does not equal 1/log_e 3; it diverges to infinity.</p><p>∴ Answer: A (False)</p>
Correct Answer: A