Question

Explain why $\underset{k\to \infty }{\lim }{k}^{\frac{1}{k}}$has an indeterminate form of the type ${\infty }^0.$Then show that this limit equals$1.$

Short answer

Limit $\underset{k\to \infty }{\lim }{k}^{\frac{1}{k}}$has an indeterminate form of the type ${\infty }^0$by substituting the limit value for k.

$\underset{k\to \infty }{\lim }{k}^{\frac{1}{k}}={\infty }^{\frac{1}{\infty }}={\infty }^0$

The value of the limit is $1$as follows.

$$\begin{gathered} \underset{k\to \infty }{\lim }{k}^{\frac{1}{k}}=\underset{k\to \infty }{\lim }{e}^{\ln \left(k^{\frac{1}{k}}\right)} \\ =\underset{k\to \infty }{\lim }{e}^{\frac{1}{k}\ln k} \\ =e^0=1 \end{gathered}$$

Step-by-step solution

Step 1. Given information.

Limit $\underset{k\to \infty }{\lim }{k}^{\frac{1}{k}}$has an indeterminate form of the type${\infty }^0.$

Step 2. Verifying the indeterminate form of the type ∞0.

Determine the value of the given limit by substituting the limit value for k.

$$\begin{gathered} \underset{k\to \infty }{\lim }{k}^{\frac{1}{k}}={\infty }^{\frac{1}{\infty }} \\ ={\infty }^0 \end{gathered}$$

so$\underset{k\to \infty }{\lim }{k}^{\frac{1}{k}}$has an indeterminate form of the type ${\infty }^0.$

Step 2. Value of limit.

Determine the value of the limit by using the logarithm property$e^{\ln x}=x.$

role="math" localid="1649194572093" $$\begin{gathered} \underset{k\to \infty }{\lim }{k}^{\frac{1}{k}}=\underset{k\to \infty }{\lim }{e}^{\ln \left(k^{\frac{1}{k}}\right)} \\ =\underset{k\to \infty }{\lim }{e}^{\frac{1}{k}\ln k} \\ =e^{\frac{1}{\infty }\ln \infty } \\ =e^0 \\ =1 \end{gathered}$$

So$\underset{k\to \infty }{\lim }{k}^{\frac{1}{k}}==1.$