Question

Calculate each of the limits in Exercises 49–64. Some of these limits are made easier by considering the logarithm of the limit first, and some are not.

$\underset{x\to 1}{\lim }{(\ln x)}^{x-1}$

Short answer

limit doesn't exist

Step-by-step solution

Step 1. Given information

$\underset{x\to 1}{\lim }{(\ln x)}^{x-1}$

Step 2. Taking log on both sides

$\begin{array}{r}\underset{x\to 1^{+}}{lim} (\ln x{)}^{x-1}=\underset{x\to 1^{+}}{lim} e^{\ln (\ln x{)}^{x-1}}\\ =\underset{x\to 1^{+}}{lim} e^{(x-1)\ln (\ln x)}\\ =e^{\underset{x\to 1^{+}}{lim} (x-1)\ln (\ln x)}\\ \\ \end{array}$

Step 3. Calculate the right limit

$$\begin{gathered} \underset{x\to 1^{+}}{lim} (\ln x{)}^{x-1}=e^0 \\ \underset{x\to 1^{+}}{lim} (\ln x{)}^{x-1}=1 \end{gathered}$$

Step 4. Calculating the left limit

$\begin{array}{r}\underset{x\to 1^{-}}{lim} (\ln x{)}^{x-1}=\underset{x\to 1^{-}}{lim} e^{\ln (\ln x{)}^{x-1}}\\ =\underset{x\to 1-}{lim} e^{(x-1)\ln (\ln x)}\\ =e^{\underset{x\to 1^{-}}{\lim }(x-1)\ln (\ln x)}\\ =e^{-\infty }\\ =\frac{1}{e^{\infty }}=0\\ \end{array}$

since $\begin{array}{r}\underset{x\to 1^{+}}{lim} (\ln x{)}^{x-1}=1\\ \underset{x\to 1^{-}}{lim} (\ln x{)}^{x-1}=0\end{array}$

both limits are different hence limit does not exist.