Q. 32 - Calculus

Question

Calculate each of the limits in Exercises $29-70$.

$\underset{x\to 1.7}{\lim } \left(3.1x^2-4x+0.8\right)$.

Step-by-Step Solution

Answer

The value of $\underset{x\to 1.7}{\lim } \left(3.1x^2-4x+0.8\right)$ is, $2.959$.

Step 1 . Given information

$\underset{x\to 1.7}{\lim } \left(3.1x^2-4x+0.8\right)$.

Step 2 . The limit is calculated as below:

$\underset{x\to 1.7}{\lim } \left(3.1x^2-4x+0.8\right)=\underset{x\to 1.7}{\lim }(3.1x^2)-\underset{x\to 1.7}{\lim }\left(4x\right)+\underset{x\to 1.7}{\lim }(0.8)$ [Sum rule]

                                       $=3.1\underset{x\to 1.7}{\lim }\left(x^2\right)-4\underset{ x\to 1.7}{\lim }\left(x\right)+0.8$ [Constant multiple rule]

                                       $$\begin{gathered} =3.1{\left(1.7\right)}^2-4\left(1.7\right)+0.8 \\ =3.1\left(2.89\right)-6.8+0.8 \\ =8.959-6.0 \\ =2.959 \end{gathered}$$