\( \lim _{x \rightarrow 0+} (\frac{x}{x-1}-\frac{1}{lnx}) \)

To solve the limit \( \lim _{x \rightarrow 0+} \left(\frac{x}{x-1}-\frac{1}{\ln x}\right) \), we first analyze the individual components as \( x \) approaches 0 from the right. 1. **Examine \( \frac{x}{x-1} \)**: As \( x \to 0^+ \), this expression simplifies to: \[ \frac{x}{x - 1} = \frac{x}{-1} = -x \to 0. \] 2. **Examine \( \frac{1}{\ln x} \)**: As \( x \to 0^+ \), \( \ln x \to -\infty \), which means: \[ \frac{1}{\ln x} \to 0. \] Putting it together, we see that both terms approach \( 0 \) as \( x \) goes to \( 0^+ \). However, we need to look at how quickly they approach \( 0 \) to evaluate their difference. To clarify this, we can rewrite the limit. We aim to combine the two fractions into a single expression: \[ \lim _{x \rightarrow 0+} \left( \frac{x}{x-1} - \frac{1}{\ln x} \right) = \lim _{x \rightarrow 0+} \left( \frac{x \ln x - (x - 1)}{(x - 1) \ln x} \right). \] Next, let's compute the limit \( x \ln x \) as \( x \to 0^+ \): We know \( \ln x \) approaches \( -\infty \) while \( x \to 0^+ \). Thus, \( x \ln x \) approaches \( 0 \) since \( x \) approaches \( 0 \) faster than \( \ln x \) approaches \( -\infty \). Now compute: \[ x \ln x - (x - 1) = x \ln x + 1 - x. \] Notice that as \( x \to 0^+ \), \( 1 - x \to 1 \) and \( x \ln x \to 0 \) gives: \[ x \ln x + 1 - x \to 1. \] Thus we are left with: \[ \lim_{x \to 0^+} \frac{1}{(-1)(-\infty)} \to 0. \] To summarize: The entire expression goes to \( \frac{1}{(x-1)(\ln x)} \) results in: \[ \lim _{x \to 0+} \left( \frac{x \ln x + 1 - x}{(x-1) \ln x} \right) = \frac{1}{0} \to 0. \] Therefore, we find: \[ \lim _{x \rightarrow 0+} \left(\frac{x}{x-1}-\frac{1}{\ln x}\right) = 0. \] Final Answer: \[ \boxed{0}. \]