Q. 20 - Calculus

Question

Express the following limit statement as delta–epsilon statement: 

$f\text{'}\left(3\right)=\underset{x\to 3}{lim} \frac{x^2-9}{x-3}$

Step-by-Step Solution

Answer

$\underset{\mathrm{\Delta }x\to 0}{lim} \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}$

Step 1. Given information

We have been given a limit statement as:

$f\text{'}\left(3\right)=\underset{x\to 3}{lim} \frac{x^2-9}{x-3}$

We have to express it as delta–epsilon statement.

Step 2. Express the limit as a delta-epsilon statement

Let, 

$$\begin{gathered} x_1=3 \\ {x}_2=x \\ {y}_1=9 \\ {y}_2=x^2 \end{gathered}$$

So,

$$\begin{gathered} \begin{array}{r}\mathrm{\Delta }x=x_2-x_1\\ =x-3\end{array} \\ \begin{array}{r}\mathrm{\Delta }y=y_2-y_1\\ =x^2-9\end{array} \end{gathered}$$

When $x\to 3$ then $x-3\to 0$ or $∆x\to 0$

Therefore, replace the values with the delta notations as follows:

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(3\right)=\underset{x\to 3}{lim} \frac{x^2-9}{x-3}\\ =\underset{\mathrm{\Delta }x\to 0}{lim} \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\end{array}$