Chapter 9.2, Problem 16E

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 16E

The value of the limit is $\frac{5}{7}$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to \infty }{\lim }\frac{5x^2-3x}{7x^2+1}$ .

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 $\frac{5}{7}$ as $x$ approaches to $\infty$ .

Therefore, the value of the limit is approximately $\frac{5}{7}$ .

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

  $5x^2-3x\to \infty$ and $7x^2+1\to \infty$

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

So, apply L'Hôpital's Rule to write

  $\underset{x\to \infty }{\lim }\left(\frac{5x^2-3x}{7x^2+1}\right)=\underset{x\to \infty }{\lim }\left(\frac{10x-3}{14x}\right)=\underset{x\to \infty }{\lim }\left(\frac{10-0}{14}\right)=\frac{5}{7}$

Therefore, the limit determined using both the methods is indeed the same.