Q. 72 - Calculus

Question

Prove each of the limit statements in Exercises 67–72. You
will have to bound $\delta$.

$\underset{x\to 3}{\lim } \frac{18}{x^2} =2$ 

Step-by-Step Solution

Answer

The given limit $\underset{x\to 3}{\lim } \frac{18}{x^2}=2$ is proved.

Step 1. Given Equation

$\underset{x\to 3}{\lim } \frac{18}{x^2}=2$

Step 2. Proving the limit

Given  $\epsilon >0$, choose  $\delta =min(1, \frac{\epsilon }{3})$, Then if, $0<|x-3|< \delta$ we have

Or 

$\epsilon >0 and \delta >0$ 

such that $\left|\frac{18-2x^2}{x^2}\right| \le \in$if  $|x-3|\le \delta$

$$\begin{gathered} =\left|\frac{18-2x^2}{x^2}\right| \le \in \\ =\frac{2}{x^2} \left|\right(9-x^2\left)\right| \le \in \\ =\frac{2}{x^2} \left|\right(3^2-x^2\left)\right| \le \in \\ =\frac{2}{x^2} \left|\right((3-x)(3+x)\left)\right| \le \in \end{gathered}$$

when x=3

$$\begin{gathered} =\frac{2|3+3|}{3^2} \left|\right(x-3\left)\right|\le \in \\ =\frac{2\left|6\right|}{9}|x-3| \le \in \\ =\frac{12}{9} |x-3| \le \in \\ =\frac{9}{3} |x-3| \le \in \\ =|x-3| \le \frac{3}{4}\in \end{gathered}$$

taking $\delta =\frac{3\in }{4}$ so that $|\frac{18}{x^2} -2| \le \in if |x-3| \le \delta$

Hence Proved.