Question:

<p><strong>Step 1:</strong> Rewrite the base: <br/>$$\frac{x}{x+1} = 1 - \frac{1}{x+1}$$</p><p><strong>Step 2:</strong> Use the standard limit form with correction terms:<br/>$$\left(1 - \frac{1}{x+1}\right)^x = e^{x\ln(1-1/(x+1))}$$</p><p><strong>Step 3:</strong> Expand the logarithm:<br/>$$\ln\left(1 - \frac{1}{x+1}\right) = -\frac{1}{x+1} - \frac{1}{2(x+1)^2} - O(1/x^3)$$</p><p><strong>Step 4:</strong> Multiply by x:<br/>$$x\ln\left(1 - \frac{1}{x+1}\right) = -\frac{x}{x+1} - \frac{x}{2(x+1)^2} - O(1/x^2)$$<br/>$$= -1 + \frac{1}{x+1} - \frac{x}{2(x+1)^2} + O(1/x^2)$$<br/>$$= -1 + \frac{1}{x} + O(1/x^2)$$</p><p><strong>Step 5:</strong> Therefore:<br/>$$\left(\frac{x}{x+1}\right)^x = e^{-1+1/x + O(1/x^2)} = \frac{1}{e} \cdot e^{1/x + O(1/x^2)} = \frac{1}{e}\left(1 + \frac{1}{x} + O(1/x^2)\right)$$</p><p><strong>Step 6:</strong> Compute the limit:<br/>$$\lim_{x\to\infty} x\left(\frac{1}{e}\left(1 + \frac{1}{x}\right) - \frac{1}{e}\right) = \lim_{x\to\infty} x \cdot \frac{1}{ex} = \frac{1}{e}$$</p><p>∴ Answer: <strong>1/e</strong></p>