Chapter 13.2, Problem 16E

To determine

To evaluate the limit if it exist

Answer to Problem 16E

The solution is $\frac{\text{'}1\text{'}}{2}$

Explanation of Solution

Given information:

Given equation is,

  $\underset{x\to 0}{\lim }\frac{\sqrt{1+h}-1}{h}$

CALCULATION:

To find the limit,

  $\underset{x\to a}{\lim f\left(x\right)}=L$

The limit of $f\left(x\right)$ as $x$ approaches $+a$ , equals $L\text{'}$ .

If the values of $f\left(x\right)$ is made arbitrarily close to $L\text{'}$ by talking $x$ to be sufficiently close to a , but not equal to a .

Given equation is,

  $\underset{x\to a}{\lim f\left(x\right)}=L$

Here can’t apply limit of a quotient law immediately, since the limit of the denominator is 0. There the preliminary algebra consist of rationalizing the numerator.

  $\begin{array}{l}\underset{x\to 0}{\lim }\frac{\sqrt{1+h}-1}{h}\\ =\underset{x\to 0}{\lim }\frac{\sqrt{1+h}-1}{h}.\frac{\sqrt{1+h}+1}{\sqrt{1+h}+1}\text{}\left[\text{rationlizing numerator}\right]\\ =\underset{x\to 0}{\lim }\frac{1+h-1}{h\left(\sqrt{1+h}+1\right)}\\ =\underset{x\to 0}{\lim }\frac{h}{h\left(\sqrt{1+h}+1\right)}\\ =\underset{x\to 0}{\lim }\frac{1}{h\left(\sqrt{1+h}+1\right)}\\ =\frac{1}{\sqrt{1+h+1}}\left[\text{substitute }h=0\right]\\ =\frac{1}{1+1}\\ =\frac{1}{2}\end{array}$

Hence,

The solution is $\frac{\text{'}1\text{'}}{2}$