55. \( \lim _{x \rightarrow 1^{+}}(\ln x)^{x-1} \)

To find the limit \( \lim _{x \rightarrow 1^{+}}(\ln x)^{x-1} \), we start by examining the behavior of \( \ln x \) as \( x \) approaches \( 1 \) from the right. As \( x \to 1^{+} \), \( \ln x \to \ln 1 = 0 \). Also, \( x - 1 \to 0 \). Therefore, we have a limit of the form \( 0^0 \), which is indeterminate. To resolve this, we can rewrite the expression using the exponential function: \[ (\ln x)^{x-1} = e^{(x-1) \ln(\ln x)} \] Now, we need to analyze the limit of the exponent: \[ \lim_{x \to 1^+} (x-1) \ln(\ln x) \] Next, we substitute \( y = x - 1 \), which means as \( x \to 1^+ \), \( y \to 0^+ \) and \( x = y + 1 \). Now, we rewrite \( \ln x \) in terms of \( y \): \[ \ln x = \ln(y + 1) \] Using the Taylor expansion around \( y = 0 \): \[ \ln(y + 1) \approx y \text{ as } y \to 0 \] Thus: \[ \ln(\ln x) \approx \ln(y) \text{ as } y \to 0 \] Now we have: \[ \lim_{y \to 0^+} y \ln(\ln(y+1)) \approx \lim_{y \to 0^+} y \ln(y) \] The limit \( y \ln(y) \) approaches \( 0 \) as \( y \to 0^{+} \). This is due to the fact that \( \ln(y) \) approaches \( -\infty \) and dominates the term \( y \) as it tends to \( 0 \). Putting it all together, we have: \[ \lim_{x \to 1^+} (x-1) \ln(\ln x) = \lim_{y \to 0^+} y \ln(y) = 0 \] Thus, we can evaluate our original limit: \[ \lim_{x \to 1^{+}} (\ln x)^{x-1} = e^{0} = 1 \] Therefore, the final result is: \[ \boxed{1} \]