Q. 53 - 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. 

$\underset{x\to 1}{\lim } x^{\frac{1}{x-1}}$

Step-by-Step Solution

Answer

$\underset{x\to 1}{\lim } x^{\frac{1}{x-1}}=e$

Step 1. Given information

$\underset{x\to 1}{\lim } x^{\frac{1}{x-1}}$

Step 2. Taking log on both sides

$$\begin{gathered} \begin{array}{r}\underset{x\to 1^{+}}{lim} \ln {\left(x\right)}^{\frac{1}{x-1}}=\underset{x\to 1^{+}}{lim} \frac{1}{(x-1)}\ln x\\ \sin ce , \log {\left(a\right)}^b=b\log \left(a\right)\\ =\underset{x\to 1^{+}}{lim} \frac{\ln x}{(x-1)} (\frac{0}{0} form)\\ =\underset{x\to 1^{+}}{lim} \frac{1}{x} ( L\text{'} hospital rule)\\ =\underset{x\to 1^{+}}{lim} \frac{1}{x}\end{array} \\ =1 \end{gathered}$$

Step 3. Calculating the value of limit

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