Calculus and Its Applications, Chapter 1, 1.2, Section 1.2, Problem 16

Determine the $\displaystyle \lim_{x \to -1} \left( 3x^5 + 4x^4 - 3x + 6 \right)$ by using the Theorem on Limits of Rational Functions.
When necessary, state that the limit does not exist.


$
\begin{equation}
\begin{aligned}
\lim_{x \to -1} \left( 3x^5 + 4x^4 - 3x + 6 \right) &= \lim_{x \to -1} 3x^5 + \lim_{x \to -1} 4x^4 - \lim_{x \to -1} 3x + \lim_{x \to -1} 6
&& \text{The limit of a difference is the difference of the limits and the limit of a sum is the sum of the limits}\\
\\
&= 3 \cdot \lim_{x \to -1} x^5 + 4 \cdot \lim_{x \to -1} x^4 - 3 \cdot \lim_{x \to -1}x + \lim_{x \to -1} 6
&& \text{The limit of a constant times a function is the constant times the limit}\\
\\
&= 3 \cdot \left( \lim_{x \to -1}x \right)^5 + 4 \cdot \left( \lim_{x \to -1}x \right)^4 - 3 \cdot \lim_{x \to -1} x + \lim_{x \to -1} 6
&& \text{The limit of a power is the power of the limit}\\
\\
&= 3 \cdot \left( \lim_{x \to -1}x \right)^5 + 4 \cdot \left( \lim_{x \to -1}x \right)^4 - 3 \cdot \lim_{x \to -1}x + 6
&& \text{The limit of a constant is the constant}\\
\\
&= 3(-1)^5 + 4(-1)^4 - 3(-1) + 6
&& \text{Substitute } -1\\
\\
&= -3 + 4 +3 + 6\\
\\
&= 10
\end{aligned}
\end{equation}
$