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.

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

Short answer

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

Step-by-step solution

Step 1. Given information

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

Step 2. Taking log on both sides

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

Step 3. Calculating the value of limit

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