Question:

<div class="solution"> <p><strong>Step 1:</strong> To evaluate the given limit \(\lim_{x \to \infty} x\left(\left(\dfrac{x}{x+1}\right)^x - \dfrac{1}{e}\right)\), let's first consider the expression \(\left(\dfrac{x}{x+1}\right)^x\). This resembles the limit form of \(e\), where \(\lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x = e\). By manipulating the given expression, we can try to bring it into a form that involves \(e\) or a known limit.</p> <p><strong>Step 2:</strong> We rewrite \(\left(\dfrac{x}{x+1}\right)^x\) as \(\left(\dfrac{1}{1+\frac{1}{x}}\right)^x\). This form is more recognizable and can be related to the limit definition of \(e\). We know that \(\lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x = e\), so \(\lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^{-x} = \dfrac{1}{e}\). Thus, \(\left(\dfrac{x}{x+1}\right)^x\) approaches \(\dfrac{1}{e}\) as \(x\) approaches infinity. To evaluate the original limit, we can consider using L'Hôpital's Rule after rewriting the expression in a suitable form.</p> <p><strong>Step 3:</strong> Let \(f(x) = \left(\dfrac{x}{x+1}\right)^x\). Taking the natural logarithm of both sides gives \(\ln(f(x)) = x \ln\left(\dfrac{x}{x+1}\right)\). By using the property of logarithms, we can simplify this to \(\ln(f(x)) = x \ln\left(\dfrac{1}{1+\frac{1}{x}}\right)\). Expanding the logarithm and simplifying, we aim to find a series expansion that helps in evaluating the limit. Considering the series expansion of \(\ln(1 + x)\) and \(\ln(1 - x)\) for small \(x\), we have \(\ln(1 + x) \approx x - \dfrac{x^2}{2}\) and \(\ln(1 - x) \approx -x - \dfrac{x^2}{2}\) for small \(x\). Applying a similar approach to our expression, we can find an approximation for \(\left(\dfrac{x}{x+1}\right)^x\) as \(x\) approaches infinity.</p> <p><strong>Step 4:</strong> To find the limit of \(x\left(\left(\dfrac{x}{x+1}\right)^x - \dfrac{1}{e}\right)\) as \(x\) approaches infinity, we recognize that as \(x\) becomes very large, \(\dfrac{x}{x+1}\) approaches 1. Thus, we can use the approximation \(\ln\left(\dfrac{x}{x+1}\right) \approx -\dfrac{1}{x}\) for large \(x\), since \(\ln(1 - \dfrac{1}{x}) \approx -\dfrac{1}{x} - \dfrac{1}{2x^2}\) for small \(\dfrac{1}{x}\). Using this approximation, \(x\ln\left(\dfrac{x}{x+1}\right) \approx x \cdot \left(-\dfrac{1}{x} - \dfrac{1}{2x^2}\right) = -1 - \dfrac{1}{2x}\). Therefore, \(\left(\dfrac{x}{x+1}\right)^x \approx e^{-1 - \frac{1}{2x}}\). Expanding \(e^{-1 - \frac{1}{2x}}\) gives us \(\dfrac{1}{e} \cdot e^{-\frac{1}{2x}} \approx \dfrac{1}{e} \cdot \left(1 - \dfrac{1}{2x}\right)\) for large \(x\), using the approximation \(e^x \approx 1 + x\) for small \(x\). Substituting this back into the original expression, we get \(x\left(\dfrac{1}{e} \cdot \left(1 - \dfrac{1}{2x}\