(6) \( \lim _{x \rightarrow 1} \frac{\tan \pi x}{x-1} \)

To solve the limit \( \lim _{x \rightarrow 1} \frac{\tan \pi x}{x-1} \), we first substitute \( x = 1 \): \[ \text{The numerator becomes } \tan(\pi \cdot 1) = \tan(\pi) = 0. \] \[ \text{The denominator becomes } 1 - 1 = 0. \] This results in a \( \frac{0}{0} \) indeterminate form, so we can apply L'Hôpital's Rule. According to L'Hôpital's Rule, we need to differentiate the numerator and denominator: 1. The derivative of the numerator \( \tan(\pi x) \) is \( \pi \sec^2(\pi x) \) (using the chain rule). 2. The derivative of the denominator \( x - 1 \) is simply \( 1 \). Now, we apply L'Hôpital's Rule: \[ \lim _{x \rightarrow 1} \frac{\tan \pi x}{x-1} = \lim _{x \rightarrow 1} \frac{\pi \sec^2(\pi x)}{1}. \] Next, we evaluate at \( x = 1 \): \[ \sec^2(\pi \cdot 1) = \sec^2(\pi) = \left(\frac{1}{\cos(\pi)}\right)^2 = \left(\frac{1}{-1}\right)^2 = 1. \] Therefore, we have: \[ \lim _{x \rightarrow 1} \frac{\tan \pi x}{x-1} = \pi \cdot 1 = \pi. \] Thus, the final answer is: \[ \lim_{x \rightarrow 1} \frac{\tan \pi x}{x - 1} = \pi. \]