Chapter 14, Problem 1RE

To determine

To find:

The value of function $f\left(x\right)=\underset{x\to 2}{\lim }\left(3x^2-2x+1\right)$

Answer to Problem 1RE

  $\underset{x\to 2}{\lim }f\left(x\right)=9$

Explanation of Solution

Given:

The function $f$ is

  $f\left(x\right)=\underset{x\to 2}{\lim }\left(3x^2-2x+1\right)$

Concept used:

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$ , except possibly at $x=a$

Then

  $\underset{x\to a}{\lim }f\left(x\right)=L$

Calculation:

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$ , except possibly at $x=a$

  $\underset{x\to a}{\lim }f\left(x\right)=L$

If for every number $\in >0$ there is some number $\delta >0$ such that

  $\left|f\left(x\right)-L\right|<\in \text{when}\text{}\text{ever}0<\left|x-a\right|<\delta$

Other definition of limit

Let $f\left(x\right)$ be a function defined on an interval that contains $x=a$

  $\begin{array}{l}\underset{x\to 2}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }\left(3x^2-2x+1\right)\\ \text{apply}\text{by}\text{limit}\\ \underset{x\to 2}{\lim }f\left(x\right)=3\times {\left(2\right)}^2-2\left(2\right)+1\\ \underset{x\to 2}{\lim }f\left(x\right)=12-4+1\\ \underset{x\to 2}{\lim }f\left(x\right)=9\end{array}$