Chapter 9.2, Problem 26E

To determine

To Find: The value of a limit using L'Hôpital's Rule, after identifying the indeterminate form.

Answer to Problem 26E

The value of the limit is $1$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to \infty }{\lim }{\left(\ln x\right)}^{\frac{1}{x}}$ .

Method: $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(x)\text{ and }g(x)\text{ approach infinity as }x\to c,\text{ then }\underset{x\to c}{\lim }\frac{f(x)}{g(x)}=\underset{x\to c}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: As $a\to \infty$ ,

  $\ln a\to \infty$ and $\frac{1}{a}\to 0$

Therefore, the limit is of the indeterminate form ${\infty }^0$ .

So, assume the limit is $L$ and take the natural logarithm of the equation $L=\underset{x\to \infty }{\lim }{\left(\ln x\right)}^{\frac{1}{x}}$ so that

  $\begin{array}{l}L=\underset{x\to \infty }{\lim }{\left(\ln x\right)}^{\frac{1}{x}}\\ \Rightarrow \ln L=\ln \left[\underset{x\to \infty }{\lim }{\left(\ln x\right)}^{\frac{1}{x}}\right]\\ \Rightarrow \ln L=\underset{x\to \infty }{\lim }\left[\ln \left({\left[\ln x\right]}^{\frac{1}{x}}\right)\right]\\ \Rightarrow \ln L=\underset{x\to \infty }{\lim }\left[\frac{\ln \left(\ln x\right)}{x}\right]\end{array}$

Now, the righthand side of the above equation is in $\frac{\infty }{\infty }$ form.

So, apply L'Hôpital's Rule on the righthand side of the equation to write

  $\begin{array}{l}\ln L=\underset{x\to \infty }{\lim }\frac{\frac{1}{\ln x}\cdot \frac{1}{x}}{1}=\underset{x\to \infty }{\lim }\left(\frac{1}{x\ln x}\right)=0\\ \Rightarrow L=e^0=1\end{array}$

Therefore, the value of the limit is $1$ .