Chapter 12.1, Problem 47E

To determine

To find: Find the limit by direct substitution.

Answer to Problem 47E

The value of limit is $\underset{x\to -3}{\lim }\left(2x^2+4x+1\right)=7$

Explanation of Solution

Given information:The given limit is $\underset{x\to -3}{\lim }\left(2x^2+4x+1\right)$

The given limit is $\underset{x\to -3}{\lim }\left(2x^2+4x+1\right)$

To evaluate the limit we directly substitute the value $x=4$ in the function.

So we get

  $\begin{array}{c}\underset{x\to -3}{\lim }\left(2x^2+4x+1\right)=\left\{2{\left(-3\right)}^2+4\left(-3\right)+1\right\}\\ =\left(2\times 9-12+1\right)\\ =\left(18-12+1\right)\\ =7\end{array}$