Question

find the limit.
lim (3x^3 - 4x^2 + 3x + 1)
x→ - 4

select the correct choice below and, if necessary, fill in the answer box to complete your choice

a. lim (3x^3 - 4x^2 + 3x + 1) = (simplify your answer.)
x→ - 4

b. the limit does not exist.

Explanation:

Step1: Substitute x = - 4

Substitute \(x=-4\) into \(3x^{3}-4x^{2}+3x + 1\).
\[

$$\begin{align*} &3(-4)^{3}-4(-4)^{2}+3(-4)+1\\ \end{align*}$$

\]

Step2: Calculate \((-4)^{3}\) and \((-4)^{2}\)

We know that \((-4)^{3}=(-4)\times(-4)\times(-4)=-64\) and \((-4)^{2}=(-4)\times(-4) = 16\).
\[

$$\begin{align*} &3\times(-64)-4\times16+3\times(-4)+1\\ &=-192 - 64-12 + 1 \end{align*}$$

\]

Step3: Perform addition and subtraction

\[

$$\begin{align*} &-192-64-12 + 1\\ &=-(192 + 64+12)+1\\ &=-268 + 1\\ &=-267 \end{align*}$$

\]

Answer:

A. \(\lim_{x
ightarrow - 4}(3x^{3}-4x^{2}+3x + 1)=-267\)