Use l'Hôpital's rule to find the following limit.
\[
\lim _{x \rightarrow 1^{+}}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)
\]

Step 1 ：We are given the limit \(\lim _{x \rightarrow 1^{+}}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)\).

Step 2 ：As x approaches 1, both the numerator and the denominator approach 0. Therefore, this limit is of the form \(\frac{0}{0}\).

Step 3 ：We can apply l'Hôpital's rule, which states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

Step 4 ：First, we find the derivatives of the functions \(f(x) = \frac{1}{\ln x}\) and \(g(x) = \frac{1}{x-1}\).

Step 5 ：The derivative of \(f(x) = \frac{1}{\ln x}\) is \(f'(x) = -\frac{1}{x(\ln x)^2}\).

Step 6 ：The derivative of \(g(x) = \frac{1}{x-1}\) is \(g'(x) = -\frac{1}{(x - 1)^2}\).

Step 7 ：Applying l'Hôpital's rule, we find that the limit is equal to \(\lim _{x \rightarrow 1^{+}}\left(\frac{f'(x)}{g'(x)}\right)\).

Step 8 ：Substituting the derivatives we found into the limit, we get \(\lim _{x \rightarrow 1^{+}}\left(\frac{-\frac{1}{x(\ln x)^2}}{-\frac{1}{(x - 1)^2}}\right)\).

Step 9 ：Simplifying the expression, we find that the limit is equal to 1.

Step 10 ：Thus, the limit of the given function as x approaches 1 from the right is \(\boxed{1}\).