Chapter 9.2, Problem 27E

(a).

To determine

To Estimate: The value of a given limit, using tables.

Answer to Problem 27E

The estimated value of $\underset{x\to \infty }{\lim }\frac{\ln x^5}{x}$ is $0$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to \infty }{\lim }\frac{\ln x^5}{x}$

Method Used: Calculate the value of the expression for the given values of $x$ in the table and observe the trend.

Calculation:The values of the expression for the given values of $x$ are

  $\begin{array}{cc}x& \frac{ln{x}^5}{x}\\ \text{10}& \text{1}\text{.1513}\\ {\text{10}}^2& \text{0}\text{.2302}\\ {10}^3& \text{0}\text{.0345}\\ {10}^4& \text{0}\text{.0046}\\ {\text{10}}^5& \text{0}\text{.0005}\end{array}$

In the above table, as $x$ becomes large, the value of the expression approaches $0$ .

Therefore, the estimated value of the limit is $0$ .

(b).

To determine

To Find: The value of a limit using L'Hôpital's Rule and confirm the estimate from the previous sub-part.

Answer to Problem 27E

The value of the limit is $0$

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to \infty }{\lim }\frac{\ln x^5}{x}$

Concept Used:

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

Calculation: Apply L'Hôpital's Rule to write

  $\underset{x\to \infty }{\lim }\frac{\ln x^5}{x}=\underset{x\to \infty }{\lim }\frac{\raisebox{1ex}{$5x^4$}\!\left/ \!\raisebox{-1ex}{$x^5$}\right.}{1}$

  $=5\underset{x\to \infty }{\lim }\frac{1}{x}$

  $=5\cdot 0$

  $=0$

Therefore, the limit is $0$ and it is the same value as estimated from the previous sub-part.