To solve the given limit:

$\lim_{{x \to 1}} \left( \frac{x}{x - 1} - \frac{1}{\ln x} \right),$

we need to determine if we can apply L'Hospital's Rule, which is typically used for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Let's analyze each term individually as $x \to 1$.

Step 1: Analyzing $\frac{x}{x - 1}$

As $x \to 1$, the denominator $(x - 1)$ approaches zero, and the numerator $x$ approaches 1. This part alone does not yield a simple indeterminate form; however, $\frac{x}{x - 1}$ tends to infinity as $x$ approaches 1.

Step 2: Analyzing $\frac{1}{\ln x}$

As $x \to 1$, $\ln x$ approaches zero from below (since $\ln(1) = 0$), making $\frac{1}{\ln x}$ tend to $-\infty$.

Step 3: Combining Terms

Now, we have two terms: $\frac{x}{x - 1} \to \infty \quad \text{and} \quad \frac{1}{\ln x} \to -\infty.$ This creates an $\infty - \infty$ indeterminate form, suggesting that L'Hospital's Rule might be useful.

Step 4: Rewriting and Applying L'Hospital's Rule

Rewrite the expression as a single fraction: $\lim_{{x \to 1}} \left( \frac{x}{x - 1} - \frac{1}{\ln x} \right) = \lim_{{x \to 1}} \frac{x \ln x - (x - 1)}{(x - 1) \ln x}.$

Now, we can try to apply L'Hospital's Rule by differentiating the numerator and denominator with respect to $x$. Differentiating:

1. Numerator: The derivative of $x \ln x - (x - 1)$ is $\ln x + 1 - 1 = \ln x$.
2. Denominator: The derivative of $(x - 1) \ln x$ is (\ln x + \frac{x - 1}{x} = \ln x + \frac{1 - 1}{1} = \ln