Chapter 9.2, Problem 8E

To determine

To Find: The value of a limit using L'Hôpital's Rule

Answer to Problem 8E

The value of the limit is $\frac{1}{6}$

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 2}{\lim }\frac{x^2-4x+4}{x^3-12x+16}$

Concept Used:

  $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(a)=g(a)=0,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: At $x=2$ , the value of the expression is $\frac{2^2-4\cdot 2+4}{2^3-12\cdot 2+16}=\frac{4-8+4}{8-24+16}=\frac{0}{0}$

So, apply L'Hôpital's Rule to write

  $\begin{array}{l}\underset{x\to 2}{\lim }\frac{x^2-4x+4}{x^3-12x+16}=\underset{x\to 2}{\lim }\frac{2x-4+0}{3x^2-12+0}\\ =\underset{x\to 2}{\lim }\frac{2x-4}{3x^2-12}\end{array}$

Again, when $x=2$ , the value of $\frac{2x-4}{3x^2-12}$ is $\frac{2\times 2-4}{2\times 2^2-12}=\frac{4-4}{12-12}=\frac{0}{0}$

Therefore, apply L'Hôpital's Rule again to write

  $\begin{array}{l}\underset{x\to 2}{\lim }\frac{2x-4}{3x^2-12}=\underset{x\to 2}{\lim }\frac{2}{6x}\\ =\frac{2}{2\cdot 6}\\ =\frac{1}{6}\end{array}$

Therefore, the limit is $\frac{1}{6}$