Chapter 2.3, Problem 26E

(a)

To determine

To estimate: The value of the function to two decimal places when x approaches zero by using the graph of the function $f\left(x\right)$.

Answer to Problem 26E

The estimated value of the function when x approaches zero $f\left(x\right)$ approaches to 0.29.

Explanation of Solution

Given:

The function $f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}$.

Use the online graphing calculator to draw the graph of the function $f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}$.

From the graph, as $x=0$, then $f\left(x\right)$ is not defined. But x approaches 0, and then $f\left(x\right)$ tends to 0.289.

That is, $\underset{x\to 0}{\lim }\frac{\sqrt{3+x}-\sqrt{3}}{x}\approx 0.29$.

Thus, the estimated value of $\underset{x\to 0}{\lim }\frac{\sqrt{3+x}-\sqrt{3}}{x}$ is $\underset{\_}{0.29}$.

(b)

To determine

To estimate: The value of the limit to four decimal places when x close to 0 by using the table of values of $f\left(x\right)$.

Answer to Problem 26E

The estimated value of the limit by using the table of values of $f\left(x\right)$ for x close to 0 is $\underset{\_}{0.2887}$.

Explanation of Solution

Calculation:

Construct the table of values of $f\left(x\right)$ for x close to 0.

x	$f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}$
−0.001	$\frac{\sqrt{3+\left(-0.001\right)}-\sqrt{3}}{-0.001}\approx 0.2886992$
−0.000 1	$\frac{\sqrt{3+\left(-0.0001\right)}-\sqrt{3}}{-0.0001}\approx 0.2886775$
−0.000 01	$\frac{\sqrt{3+\left(-0.00001\right)}-\sqrt{3}}{-0.00001}\approx 0.2886754$
−0.000 001	$\frac{\sqrt{3+\left(-0.000001\right)}-\sqrt{3}}{-0.000001}\approx 0.2886752$
0.000 001	$\frac{\sqrt{3+\left(0.000001\right)}-\sqrt{3}}{0.000001}\approx 0.2886751$
0.000 01	$\frac{\sqrt{3+\left(0.00001\right)}-\sqrt{3}}{0.00001}\approx 0.2886749$
0.000 1	$\frac{\sqrt{3+\left(0.0001\right)}-\sqrt{3}}{0.0001}\approx 0.2886727$
0.001	$\frac{\sqrt{3+\left(0.001\right)}-\sqrt{3}}{0.001}\approx 0.2886511$

From the table, $f\left(x\right)$ approaches 0.2887 as x gets more close to 0.

That is, $\underset{x\to 0}{\lim }\frac{\sqrt{3+x}-\sqrt{3}}{x}\approx 0.2887$.

Thus, the limit appears to be approximately equal to $\underset{\_}{0.2887}$.

(c)

To determine

To find: The exact limit value of the function $f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}$ as x approaches 0.

Answer to Problem 26E

The exact limit value of the function is $\underset{\_}{\frac{1}{2\sqrt{3}}}$.

Explanation of Solution

Given:

The limit of the function as x approaches 0 is $f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}$.

Limit Laws:

Suppose that c is a constant and the limits $\underset{x\to a}{\lim }f\left(x\right)$ and $\underset{x\to a}{\lim }g\left(x\right)$ exist, then

Limit law 1: $\underset{x\to a}{\lim }\left[f\left(x\right)+g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }g\left(x\right)$

Limit law 2: $\underset{x\to a}{\lim }\left[f\left(x\right)-g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)-\underset{x\to a}{\lim }g\left(x\right)$

Limit law 3: $\underset{x\to a}{\lim }\left[cf\left(x\right)\right]=c\underset{x\to a}{\lim }f\left(x\right)$

Limit law 4: $\underset{x\to a}{\lim }\left[f\left(x\right)g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)\cdot \underset{x\to a}{\lim }g\left(x\right)$

Limit law 5: $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)}$ if $\underset{x\to a}{\lim }g\left(x\right)\ne 0$

Limit law 7: $\underset{x\to a}{\lim }c=c$

Limit law 8: $\underset{x\to a}{\lim }x=a$

Limit law 11: $\underset{x\to a}{\lim }\sqrt[n]{f\left(x\right)}=\sqrt[n]{\underset{x\to a}{\lim }f\left(x\right)}$ where n is a positive integer, if n is even, assume that $\underset{x\to a}{\lim }f\left(x\right)>0$.

Direct substitution property:

If f is a polynomial or a rational function and a is in the domain of f, then $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$.

Difference of square formula: $\left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)$

Fact 1:

If $f\left(x\right)=g\left(x\right)$ when $x\ne a$, then $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }g\left(x\right)$, provided the limit exist.

Let $f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}$ (1)

Note 1:

The direct substitution method is not applicable for the function $f\left(x\right)$ as the function $f\left(0\right)$ is in indeterminate form when $x=0$.

$\begin{array}{c}f\left(0\right)=\frac{\sqrt{3+0}-\sqrt{3}}{0}\\ =\frac{\sqrt{3}-\sqrt{3}}{0}\\ =\frac{0}{0}\end{array}$

Note 2:

The Quotient rule is not applicable for the function $f\left(x\right)$ as the limit of the denominator is zero.

$\underset{x\to 0}{\lim }\left(x\right)=0\left[\text{ by limit law 8}\right]$

Note 3:

“The limit may be infinite or some finite value when both the numerator and the denominator approach 0.”

Calculation:

By note 3, take the limit x approaches 0 but $x\ne 0$.

Simplify $f\left(x\right)$ by using elementary algebra, $f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}$.

Take the conjugate of the numerator and multiply and divide by $f\left(x\right)$.

$\begin{array}{c}f\left(x\right)=\frac{\sqrt{3+x}-\sqrt{3}}{x}\times \frac{\sqrt{3+x}+\sqrt{3}}{\sqrt{3+x}+\sqrt{3}}\\ =\frac{\left(\sqrt{3+x}-\sqrt{3}\right)\left(\sqrt{3+x}+\sqrt{3}\right)}{x\left(\sqrt{3+x}+\sqrt{3}\right)}\end{array}$

Apply the formula for the difference of square,

$\begin{array}{c}f\left(x\right)=\frac{\left({\left(\sqrt{3+x}\right)}^2-{\left(\sqrt{3}\right)}^2\right)}{x\left(\sqrt{3+x}+\sqrt{3}\right)}\\ =\frac{\left(3+x-3\right)}{x\left(\sqrt{3+x}+\sqrt{3}\right)}\\ =\frac{x}{x\left(\sqrt{3+x}+\sqrt{3}\right)}\end{array}$

Since the limit x approaches 0 but not equal to 0, cancel the common term $x\ne 0$ from both the numerator and the denominator,

$f\left(x\right)=\frac{1}{\left(\sqrt{3+x}+\sqrt{3}\right)}$

Use fact 1, $f\left(x\right)=\frac{1}{\left(\sqrt{3+x}+\sqrt{3}\right)}$ and $x\ne 0$, then $\underset{x\to 0}{\lim }\frac{\sqrt{3+x}-\sqrt{3}}{x}=\underset{x\to 0}{\lim }\frac{1}{\left(\sqrt{3+x}+\sqrt{3}\right)}$.

Use the limit laws to find the required limit function.

$\begin{array}{c}\underset{x\to 0}{\lim }\frac{1}{\left(\sqrt{3+x}+\sqrt{3}\right)}=\frac{\underset{x\to 0}{\lim }1}{\underset{x\to 0}{\lim }\left(\sqrt{3+x}+\sqrt{3}\right)}\left[\text{by limit law 5}\right]\\ =\frac{\underset{x\to 0}{\lim }1}{\underset{x\to 0}{\lim }\left(\sqrt{3+x}\right)+\underset{x\to 0}{\lim }\sqrt{3}}\left[\text{by limit law 1}\right]\\ =\frac{1}{\underset{x\to 0}{\lim }\left(\sqrt{3+x}\right)+\sqrt{3}}\left[\text{by limit law 7}\right]\end{array}$

$\begin{array}{c}=\frac{1}{\sqrt{\underset{x\to 0}{\lim }\left(3+x\right)}+\sqrt{3}}\left[\text{by limit law 11}\right]\\ =\frac{1}{\sqrt{\underset{x\to 0}{\lim }3+\underset{x\to 0}{\lim }x}+\sqrt{3}}\left[\text{by limit law 1}\right]\\ =\frac{1}{\sqrt{3+\left(0\right)}+\sqrt{3}}\left[\text{by limit laws 7 and 8}\right]\\ =\frac{1}{2\sqrt{3}}\end{array}$

Thus, the exact limit of the function is $\underset{\_}{\frac{1}{2\sqrt{3}}}$.