Chapter 9.2, Problem 3E

To determine

To Find:The value of a limit, both graphically and using L'Hôpital's Rule.

Answer to Problem 3E

The value of the limit is $\frac{1}{4}$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 2}{\lim }\frac{\sqrt{2+x}-2}{x-2}$

Method:

1. Graphically: : Plot the graph of the expression in $x$ and identify the value to which the expression approaches as $x$ approaches to $2$ .
2. $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(a)=g(a)=0,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation:

(i)The graph of the expression is

  

In the above graph, the value of the expression approaches to $\frac{1}{4}$ near $x=2$ .

Therefore, the value of the limit is $\frac{1}{4}$ .

(ii). Apply L'Hôpital's Rule to write

  $\underset{x\to 2}{\lim }\left(\frac{\sqrt{x+2}-2}{x-2}\right)=\underset{x\to 2}{\lim }\left(\frac{\frac{1}{2\sqrt{x+2}}-0}{1-0}\right)=\underset{x\to 2}{\lim }\left(\frac{1}{2\sqrt{x+2}}\right)=\frac{1}{2\times \sqrt{2+2}}=\frac{1}{4}$

The limit determined using both the methods is indeed the same.