Question:

<div class="solution"> <p><strong>Step 1:</strong> We start by examining the given limit expression: \(\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\dfrac{e^{\frac{k}{n}}+e^{-\frac{k}{n}}}{n\sqrt{1-e^{\frac{2k}{n}}-e^{-\frac{2k}{n}}}}\). This looks like a Riemann sum, which is a method for approximating the value of a definite integral.</p> <p><strong>Step 2:</strong> To simplify the expression under the square root in the denominator, we notice that \(1 - e^{\frac{2k}{n}} - e^{-\frac{2k}{n}} = 1 - (e^{\frac{k}{n}})^2 - (e^{-\frac{k}{n}})^2 = (1 - e^{\frac{k}{n}})(1 + e^{\frac{k}{n}}) - (e^{-\frac{k}{n}})^2 = (1 - e^{\frac{k}{n}})(1 + e^{\frac{k}{n}}) - (1 - e^{\frac{k}{n}} + e^{\frac{2k}{n}}) = -2e^{\frac{k}{n}} + e^{\frac{2k}{n}} + 1 - e^{\frac{2k}{n}} = 1 - 2e^{\frac{k}{n}} + (e^{\frac{k}{n}})^2 - (e^{\frac{k}{n}})^2 = (1 - e^{\frac{k}{n}})^2 - (e^{\frac{k}{n}})^2\). However, simplifying this directly may not lead to an intuitive next step. Instead, recognizing that the denominator involves terms that resemble the formula for the hyperbolic cosine and its relation to exponentials, let's consider simplifying the expression by using the identity for the hyperbolic cosine: \(\cosh(x) = \frac{e^x + e^{-x}}{2}\) and the fact that \(1 - \cosh(2x) = -\sinh^2(x)\) where \(\sinh(x) = \frac{e^x - e^{-x}}{2}\). Applying these identities, the expression can be simplified by recognizing \(e^{\frac{k}{n}} + e^{-\frac{k}{n}} = 2\cosh\left(\frac{k}{n}\right)\) and \(1 - e^{\frac{2k}{n}} - e^{-\frac{2k}{n}} = 1 - 2\cosh\left(\frac{2k}{n}\right) = -2\sinh^2\left(\frac{k}{n}\right)\), thus the denominator becomes \(n\sqrt{-2\sinh^2\left(\frac{k}{n}\right)} = n\sqrt{2}\sinh\left(\frac{k}{n}\right)\).</p> <p><strong>Step 3:</strong> Substituting these simplifications back into the original expression gives us \(\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\dfrac{2\cosh\left(\frac{k}{n}\right)}{n\sqrt{2}\sinh\left(\frac{k}{n}\right)}\). Simplifying further yields \(\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\dfrac{\cosh\left(\frac{k}{n}\right)}{n\sinh\left(\frac{k}{n}\right)}\). Recognizing that as \(n \to \infty\), the sum approaches an integral, we can rewrite this as \(\int_{0}^{1} \frac{\cosh(x)}{\sinh(x)} dx\). Let \(u = \sinh(x)\), then \(du = \cosh(x) dx\), and the integral becomes \(\int \frac{1}{u} du = \ln|u| + C\). Substituting back for \(u\) gives us \(\ln|\sinh(x)|\) evaluated from 0 to 1.</p> <p><strong>Step 4:</strong> Evaluating \(\ln|\sinh(x)|\) from 0 to 1 yields \(\ln|\sinh(1)| - \ln|\sinh(0)| = \ln|\sinh(1)|\) since \(\sinh(0) = 0\) and \(\ln(0)\) approaches \(-\