Q. 59 - Calculus

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.

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

Step-by-Step Solution

Answer

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

Step 1. Given information

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

Step 2. Taking log on both sides

$\begin{array}{r}\underset{x\to 1^{+}}{lim} \ln (x-1{)}^{\ln x}=\underset{x\to 1^{+}}{lim} (\ln x)\ln (x-1)\\ =\underset{x\to 1^{+}}{lim} \frac{\ln (x-1)}{\frac{1}{\ln x}}\left[\text{ in the form of }\frac{\mathrm{\infty }}{\mathrm{\infty }}\right]\\ =\underset{x\to 1^{+}}{lim} \left(\frac{\frac{1}{x-1}}{-(\ln x{)}^{-2}\cdot \frac{1}{x}}\right)\text{ [using L'Hospitals rule] }\\ =\underset{x\to 1^{+}}{lim} \left(\frac{\frac{1}{x-1}}{-\frac{1}{x(\ln x{)}^2}}\right)\end{array}$

$\begin{array}{r}=-\underset{x\to 1^{+}}{lim} \left(\frac{1}{x-1}\cdot x(\ln x{)}^2\right)\\ =-\underset{x\to 1^{+}}{lim} \left(\frac{x(\ln x{)}^2}{x-1}\right)\left[\text{ in the form of }\frac{\mathrm{\infty }}{\mathrm{\infty }}\right]\\ =-\underset{x\to 1^{+}}{lim} \left(\frac{x\cdot 2(\ln x)\cdot \frac{1}{x}+(\ln x{)}^2\cdot 1}{1}\right)\\ =-\underset{x\to 1^{+}}{lim} \left(2(\ln x)+(\ln x{)}^2\right)\\ =-2\ln 1-(\ln 1{)}^2\\ =0-0\\ =0\end{array}$

Step 3. Calculating the value of limit

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