Chapter 9.2, Problem 15E

To determine

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

Answer to Problem 15E

The value of the limit is ${\log }_2^e\approx 0.7$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to \infty }{\lim }\frac{\ln (x+1)}{{\log }_{2}x}$ .

Method:

Graphically: : Plot the graph of the expression in $x$ and identify the value to which the expression approaches as $x$ approaches to $\infty$ . $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(x)\text{ and }g(x)\text{ approach infinity as }x\to a,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation:

(i)The graph of the expression is

  

In the above graph, the value of the expression approaches to approximately $0.7$ as $x$ approaches to $\infty$ .

Therefore, the value of the limit is approximately $0.7$ .

(ii). As $a\to \infty$ ,

  $\ln (1+a)\to \infty$ and ${\log }_{2}a\to \infty$

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

So, apply L'Hôpital's Rule to write

  $\begin{array}{l}\underset{x\to \infty }{\lim }\left(\frac{\ln (x+1)}{{\log }_{2}x}\right)=\underset{x\to \infty }{\lim }\left(\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x+1$}\right.}{{\log }_{2}e\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\right)=\ln 2\cdot \underset{x\to \infty }{\lim }\frac{x}{x+1}\\ =\ln 2\cdot \underset{x\to \infty }{\lim }\frac{1}{1+0}=\ln 2\cdot \frac{1}{1}=ln2\end{array}$

The numerical value of $\ln 2$ is $0.693\approx 0.7$ . Therefore, the limit determined using both the methods is indeed the same.