Q. 65 - Calculus

Question

Complete the proof of Theorem 7.18 by evaluating the limits of the sequences in Exercises 65 and 66.

Explain why $\underset{k\to \mathrm{\infty }}{lim} k^{1/k}$ is an indeterminate form. Use L’Hopital’s Rule or another valid method to prove that $k^{1/k}\to 1$

Step-by-Step Solution

Answer

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

Step 1. Given information

ˆk1/k → 1.

Step 2. Proof

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

Taking log on both sides $y=k^{1/k}$,

$\begin{array}{r}\ln y=\frac{\ln k}{k}\\ \underset{k\to \mathrm{\infty }}{lim} (\ln y)=\underset{k\to \mathrm{\infty }}{lim} \frac{\ln k}{k}\\ =\underset{k\to \mathrm{\infty }}{lim} \left(\frac{\frac{1}{k}}{1}\right)\\ =0\end{array}$ 

Now, 

$\underset{k\to \mathrm{\infty }}{lim} \left(y\right)=\underset{k\to \mathrm{\infty }}{lim} \left(k^{1/k}\right)$

$\begin{array}{r}\left.\underset{k\to \mathrm{\infty }}{lim} \left(y\right)=\underset{k\to \mathrm{\infty }}{lim} \left(e^{k^{v/2}}\right)\text{ (Because }\ln y=\frac{\ln k}{k}\right)\\ =e^{\underset{i\to \mathrm{\infty }}{lim} \left(k^{v/2}\right)}(\text{ Substitution) }\\ =e^0\\ =1(\text{ Simplify) }\end{array}$