Q. 8 - Calculus

Question

Use interval notation to fill in the blanks that follow. Your

answers will involve$\delta ,\epsilon , N, \mathrm{and}/\mathrm{or} M$.

If $\underset{x\to 3^{-}}{\lim }f\left(x\right)=1$ then for all $\epsilon >0$ there is some $\delta >0$ such that if $x\in \_\_\_\_\_ \mathrm{then} \in \_\_\_\_\_$.

Step-by-Step Solution

Answer

If$\underset{x\to 3^{-}}{\lim }f\left(x\right)=1$ then for all epsilon positive, there exists a delta positive such that if, $x\in (3-\delta ,3)\cup (3,3+\delta ) \mathrm{Then} f\left(x\right)\in (1-\mathrm{\epsilon },1+\mathrm{\epsilon })$.

Step 1. Given information.

The given function is $\underset{x\to 3^{-}}{\lim }f\left(x\right)=1$.

Step 2. Explanation.

From the given expression, we have, $c=3, L=1$

The limit expression can be written as a formal statement as below,

For all epsilon positive, there exists a delta positive such that if,

$x\in (3-\delta ,3)\cup (3,3+\delta ) \mathrm{Then} f\left(x\right)\in (1-\mathrm{\epsilon },1+\mathrm{\epsilon })$.

Step 3. Statement.

Therefore, statement will be written as:

If $\underset{x\to 3^{-}}{\lim }f\left(x\right)=1$ then for all epsilon positive, there exists a delta positive such that if, $x\in (3-\delta ,3)\cup (3,3+\delta ) \mathrm{Then} f\left(x\right)\in (1-\mathrm{\epsilon },1+\mathrm{\epsilon })$.