Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 7

Determine the $\lim\limits_{x \rightarrow 1} \quad \displaystyle \left(\frac{1+3x}{1+4x^2+3x^4}\right)^3$ and justify each step by indicating the appropriate limit law(s).


$
\begin{equation}
\begin{aligned}
\lim\limits_{x \rightarrow 1} \quad \displaystyle \left(\frac{1+3x}{1+4x^2+3x^4}\right)^3 &= \left(
\lim\limits_{x \rightarrow 1}
\frac{
1+3x
}
{
1+4x^2+3x^4
}
\right)^3
&& \text{(Power Law)}\\
&= \left[
\frac{
\lim\limits_{x \rightarrow 1} (1+3x)
}
{
\lim\limits_{x \rightarrow 1} (1+4x^2+3x^4)
}
\right]^3
&& \text{(Quotient Law)}\\
&= \left(
\frac{
\lim\limits_{x \rightarrow 1} 1 +
\lim\limits_{x \rightarrow 1} 3x
}
{
\lim\limits_{x \rightarrow 1} 1 +
\lim\limits_{x \rightarrow 1} 4x^2 +
\lim\limits_{x \rightarrow 1} 3x^4
}
\right)^3
&& \text{(Sum Law)}\\
&= \left(
\frac{
1 + 3 \lim\limits_{x \rightarrow 1} x
}
{
1 + 4 \lim\limits_{x \rightarrow 1} x^2 +
3 \lim\limits_{x \rightarrow 1} x^4
}
\right)^3
&& \text{(Constant Multiple and Constant Law)}\\

&= \left[
\frac{
1+3(1)
}
{
1+4(1)^2+3(1)^4
}
\right]^3
&& \text{( Special Limit Law)}\\
\\
&= \frac{1}{8}
\end{aligned}
\end{equation}\\

$