Question:

<div class="solution"> <p><strong>Step 1:</strong> To find the limit \(\displaystyle\lim_{x \to 1}\left(\frac{1}{x-1} - \frac{1}{x^a - 2x + 1}\right)\), let's first try to simplify the expression by finding a common denominator. The expression can be rewritten as \(\displaystyle\lim_{x \to 1}\left(\frac{x^a - 2x + 1 - (x-1)}{(x-1)(x^a - 2x + 1)}\right)\).</p> <p><strong>Step 2:</strong> Simplify the numerator: \(x^a - 2x + 1 - (x-1) = x^a - x - 1\). Thus, the limit becomes \(\displaystyle\lim_{x \to 1}\left(\frac{x^a - x - 1}{(x-1)(x^a - 2x + 1)}\right)\). To proceed, we should factor \(x^a - x - 1\) if possible, but since \(a\) is a positive integer, we cannot easily factor this expression without knowing more about \(a\). However, we notice that as \(x\) approaches 1, the denominator approaches 0, suggesting that we need to apply L'Hôpital's rule or find another way to simplify the expression that allows us to evaluate the limit.</p> <p><strong>Step 3:</strong> Consider the denominator \((x-1)(x^a - 2x + 1)\). As \(x\) approaches 1, \(x^a - 2x + 1\) approaches \(1 - 2 + 1 = 0\), making the denominator approach 0. For the limit to exist, the numerator must also approach 0 as \(x\) approaches 1. Since \(x^a - x - 1\) approaches \(1 - 1 - 1 = -1\) when \(x = 1\), we see a potential issue unless \(a = 2\), in which case \(x^a - x - 1 = x^2 - x - 1\), and the limit can be evaluated by factoring or using L'Hôpital's rule. Let's assume \(a = 2\) and proceed with the simplification: the expression becomes \(\displaystyle\lim_{x \to 1}\left(\frac{x^2 - x - 1}{(x-1)(x^2 - 2x + 1)}\right)\). Factoring the numerator and denominator, we get \(\displaystyle\lim_{x \to 1}\left(\frac{(x - \frac{1 + \sqrt{5}}{2})(x - \frac{1 - \sqrt{5}}{2})}{(x-1)(x-1)}\right)\), which simplifies to \(\displaystyle\lim_{x \to 1}\left(\frac{(x - \frac{1 + \sqrt{5}}{2})(x - \frac{1 - \sqrt{5}}{2})}{(x-1)^2}\right)\). Applying L'Hôpital's rule twice or simplifying further, we find that the limit exists and can be calculated directly.</p> <p><strong>Step 4:</strong> After calculating the limit with \(a = 2\), we find that \(b = \frac{3}{2}\) because \(\displaystyle\lim_{x \to 1}\left(\frac{x^2 - x - 1}{(x-1)(x^2 - 2x + 1)}\right) = \frac{3}{2}\) after simplification and application of L'Hôpital's rule. Knowing \(a = 2\) and \(b = \frac{3}{2}\), we can evaluate the given options. The expression \(\tan^{-1}(\tan(a+b)) = \tan^{-1}(\tan(2 + \frac{3}{2})) = \tan^{-1}(\tan(\frac{7}{2}))\). Since \(\frac{7}{2}\) is between \(2\pi\) and \(3\pi\), and \(\tan(\theta)\) is periodic with period \(\pi\), \(\tan^{-1}(\tan(\frac{7}{2})) = \frac{7}{2} - 2\pi\), which does not directly match any of the provided options. However, we must carefully consider each option with the calculated values of \(a\) and \(b\).</p> <p><strong>Step