Chapter 15.1, Problem 20E

To determine

To find:

The value of given limit at thatpoint .

Answer to Problem 20E

  $\underset{n\to 0}{\lim }\frac{2n^2}{n}=0$

Explanation of Solution

Given:

The function $f$ is

  $\underset{n\to 0}{\lim }\frac{2n^2}{n}$

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:

The existence of the limit

  $\begin{array}{l}\underset{n\to 0}{\lim }\frac{2n^2}{n}\\ \underset{n\to 0}{\lim }2n\end{array}$

Apply by limit

  $\begin{array}{l}=\underset{n\to 0}{\lim }2n\\ =2\left(0\right)\\ =0\end{array}$

The value of $\underset{n\to 0}{\lim }\frac{2n^2}{n}=0$ .