Chapter 2, Problem 12RE

To determine

To evaluate: The limit of the function $\underset{x\to 3}{\lim }\frac{\sqrt{x+6}-x}{x^3-3x^2}$.

Answer to Problem 12RE

The limit of the function is $\underset{\_}{-\frac{5}{54}}$.

Explanation of Solution

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 squared 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.

Evaluation:

Let $f\left(x\right)=\frac{\sqrt{x+6}-x}{x^3-3x^2}$ (1)

Note 1:

The direct substitution method is not applicable for the function $f\left(x\right)$ as the function $f\left(3\right)$ is in an indeterminate form when $x=3$.

$\begin{array}{c}f\left(3\right)=\frac{\sqrt{3+6}-3}{{\left(3\right)}^3-3{\left(3\right)}^2}\\ =\frac{\sqrt{9}-3}{27-3\left(9\right)}\\ =\frac{3-3}{27-27}\\ =\frac{0}{0}\end{array}$

Note 2:

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

Calculation:

By note 2, consider the limit x approaches to 3 but $x\ne 3$.

Simplify $f\left(x\right)$ by using elementary algebra.

$\begin{array}{c}f\left(x\right)=\frac{\sqrt{x+6}-x}{x^3-3x^2}\\ =\frac{\sqrt{x+6}-x}{x^2\left(x-3\right)}\end{array}$

Multiply both the numerator and the denominator by the conjugate of the numerator,

$f\left(x\right)=\frac{\sqrt{x+6}-x}{x^3-3x^2}\times \frac{\sqrt{x+6}+x}{\sqrt{x+6}+x}$

Apply the difference of squared formula,

$\begin{array}{c}f\left(x\right)=\frac{{\left(\sqrt{x+6}\right)}^2-{\left(x\right)}^2}{x^2\left(x-3\right)\left(\sqrt{x+6}+x\right)}\\ =\frac{x+6-x^2}{x^2\left(x-3\right)\left(\sqrt{x+6}+x\right)}\end{array}$

$=\frac{-\left(x^2-x-6\right)}{x^2\left(x-3\right)\left(\sqrt{x+6}+x\right)}$ (2)

Factorize the numerator,

$\begin{array}{c}{x}^2-x-6=x^2-3x+2x-6\\ =x\left(x-3\right)+2\left(x-3\right)\\ =\left(x-3\right)\left(x+2\right)\end{array}$

Substitute $\left(x-3\right)\left(x+2\right)$ for $\left(x^2-x-6\right)$ in equation (2),

$f\left(x\right)=\frac{-\left(x-3\right)\left(x+2\right)}{x^2\left(x-3\right)\left(\sqrt{x+6}+x\right)}$

Since the limit of x approaches 3 but not equal to 3, cancel the common term $x-3\ne 0$ from both the numerator and the denominator,

$f\left(x\right)=\frac{-\left(x+2\right)}{x^2\left(\sqrt{x+6}+x\right)}$.

Use fact 1, $f\left(x\right)=\frac{-\left(x+2\right)}{x^2\left(\sqrt{x+6}+x\right)}$ and $x\ne 3$, then $\underset{x\to 3}{\lim }\frac{\sqrt{x+6}-x}{x^3-3x^2}=\underset{x\to 3}{\lim }\frac{-\left(x+2\right)}{x^2\left(\sqrt{x+6}+x\right)}$.

Apply the direct substitution property on the limit function.

$\begin{array}{c}\underset{x\to 3}{\lim }\frac{-\left(x+2\right)}{x^2\left(\sqrt{x+6}+x\right)}=\frac{-\left(3+2\right)}{{\left(3\right)}^2\left(\sqrt{3+6}+3\right)}\\ =\frac{-\left(5\right)}{9\left(\sqrt{9}+3\right)}\\ =\frac{-\left(5\right)}{9\left(3+3\right)}\\ =-\frac{5}{54}\end{array}$

Thus, the limit of the function is $\underset{\_}{-\frac{5}{54}}$.