\( \lim _{x \rightarrow 1} (\frac{x}{x-1}-\frac{1}{lnx}) \)

To evaluate the limit \( \lim _{x \rightarrow 1} \left(\frac{x}{x-1}-\frac{1}{\ln x}\right) \), we start by noticing that both terms approach infinity as \( x \) approaches 1. Hence, we can rewrite the expression with a common denominator: \[ \frac{x \ln x - (x - 1)}{(x - 1) \ln x} \] Now, we can simplify the numerator: \[ x \ln x - x + 1 \] Next, to handle the limit, we can apply L'Hôpital's rule since we have an indeterminate form \( \frac{0}{0} \) as \( x \rightarrow 1 \). We differentiate the numerator and the denominator separately. For the numerator \( x \ln x - x + 1 \): - The derivative is \( \ln x + 1 - 1 = \ln x \). For the denominator \( (x - 1) \ln x \): - Using the product rule: \( (x - 1) \cdot \frac{1}{x} + \ln x \). Applying L'Hôpital's Rule, we compute the limit as \( x \to 1 \): \[ \lim_{x \to 1} \frac{\ln x}{(x - 1) \cdot \frac{1}{x} + \ln x} \] At \( x = 1 \), \( \ln 1 = 0 \), which gives us the \( \frac{0}{0} \) form again. We apply L'Hôpital's rule a second time. For the numerator: - The derivative of \( \ln x \) is \( \frac{1}{x} \). For the denominator: - Derivative of \( (x - 1) \cdot \frac{1}{x} + \ln x \): - The derivative of the first part is \( \frac{1}{x} - \frac{x-1}{x^2} \) and \( \frac{1}{x} \). - Combining yields: \( \frac{1}{x} - \frac{1}{x} + \frac{1}{x^2} = \frac{1}{x^2} \). So, applying L'Hôpital's rule again, we have: \[ \lim_{x \to 1} \frac{\frac{1}{x}}{\frac{1}{x^2}} = \lim_{x \to 1} x = 1. \] Therefore, the limit is \[ \boxed{1}. \]