<p>\(\lim_{x \to \infty} \left(\frac{x+6}{x+1}\right)^{x+4} =\) ______</p>
Step-by-Step Solution
Key Concept: Rewrite the fraction as 1 + (a small quantity) and use the standard limit form lim(1 + 1/n)^n = e. The exponent must be matched with the denominator of the fractional part.
<p><strong>Step 1:</strong> Recognize the indeterminate form 1^∞. Rewrite the base:</p><p>$$\frac{x+6}{x+1} = \frac{(x+1)+5}{x+1} = 1 + \frac{5}{x+1}$$</p><p><strong>Step 2:</strong> Substitute into the limit:</p><p>$$\lim_{x \to \infty} \left(1 + \frac{5}{x+1}\right)^{x+4}$$</p><p><strong>Step 3:</strong> Rewrite the exponent to match the standard form. Note that:</p><p>$$\left(1 + \frac{5}{x+1}\right)^{x+4} = \left[\left(1 + \frac{5}{x+1}\right)^{\frac{x+1}{5}}\right]^{\frac{5(x+4)}{x+1}}$$</p><p><strong>Step 4:</strong> As x → ∞, the inner bracket approaches e using the standard limit lim(1 + 1/n)^n = e:</p><p>$$\lim_{x \to \infty} \left(1 + \frac{5}{x+1}\right)^{\frac{x+1}{5}} = e$$</p><p><strong>Step 5:</strong> Evaluate the exponent:</p><p>$$\lim_{x \to \infty} \frac{5(x+4)}{x+1} = \lim_{x \to \infty} \frac{5x+20}{x+1} = 5$$</p><p><strong>Step 6:</strong> Combine results:</p><p>$$\lim_{x \to \infty} \left(\frac{x+6}{x+1}\right)^{x+4} = e^5$$</p>
Correct Answer: e^5