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

The required limit is e.

Step-by-step solution

Step 1. Given information.

Consider the given information,

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

Step 2. Taking log on both sides.

Apply the logarithm property to simplify.

$\begin{array}{rcl}\underset{x\to 1^{+}}{lim} \ln {\left(x\right)}^{\frac{1}{x-1}}& =& \underset{x\to 1^{+}}{lim} \frac{1}{(x-1)}\ln x\left[\mathrm{since},log{\left(a\right)}^b=blog\left(a\right)\right]\\ & =& \underset{x\to 1^{+}}{lim} \frac{\ln x}{(x-1)}(\frac{0}{0}\mathrm{form})\\ & =& \underset{x\to 1^{+}}{lim} \frac{1}{x}(\mathrm{L}\text{'}\mathrm{hospital}\mathrm{rule})\\ & =& \underset{x\to 1^{+}}{lim} \frac{1}{x}\\ & =& 1\end{array}$

Step 3. Calculating the value of the limit.

Find the limit from the above concepts,

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