Question

Some Convergent Sequences Involving Exponents: For any real number p > 0, the following sequences converge. Fill in each blank with the appropriate value.

$k^{\frac{1}{k}}\to$

Short answer

The required answer is$k^{\frac{1}{k}}\to 1$

Step-by-step solution

Step 1. Given information    

The given data is for any real number p > 0, the sequences converge.

Step 2. Explanation   

The given function is in indeterminate form as it is of the form ${\infty }^0$

We will solve the function by taking logarithm on both the sides,

$$\begin{gathered} y=k^{\frac{1}{k}} \\ \ln y=\frac{1}{k}\ln k \\ \underset{k\to \infty }{lim}\ln y=\underset{k\to \infty }{lim}\frac{1}{k}\ln k \\ \underset{k\to \infty }{lim}\ln y=\underset{k\to \infty }{lim}\left(\frac{{\displaystyle \frac{1}{k}}}{1}\right)(u\sin gL\text{'}Hopitalrule) \\ \underset{k\to \infty }{lim}\ln y=0 \end{gathered}$$

Further,

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