Q. 19 - Calculus

Question

Express each of the following limit statements as delta–epsilon statements:

$f\text{'}\left(3\right)=\underset{h\to 0}{\lim }\frac{{\left(3+h\right)}^2-9}{h}$

Step-by-Step Solution

Answer

$\underset{∆x\to 0}{\lim }\frac{∆y}{∆x}$

Step 1. Given information

We have been given a limit statement as:

$f\text{'}\left(3\right)=\underset{h\to 0}{\lim }\frac{{\left(3+h\right)}^2-9}{h}$

We have to express it as delta–epsilon statement.

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

Let,

$$\begin{gathered} x_1=0 \\ {x}_2=h \\ {y}_1=9 \\ {y}_2={\left(3+h\right)}^2 \end{gathered}$$

So,

$$\begin{gathered} ∆x=x_2-x_1 \\ =h-0 \\ =h \\ ∆y=y_2-y_1 \\ ={\left(3+h\right)}^2-9 \end{gathered}$$

When $h\to 0$ then $∆x\to 0$

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

$$\begin{gathered} f\text{'}\left(3\right)=\underset{h\to 0}{\lim }\frac{{\left(3+h\right)}^2-9}{h} \\ =\underset{∆x\to 0}{\lim }\frac{∆y}{∆x} \end{gathered}$$