Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 4

Determine the $\lim\limits_{x \rightarrow 2} \displaystyle \frac{2x^2+1}{x^2+6x-4}$ and justify each step by indicating the appropriate limit law(s).


$
\begin{equation}
\begin{aligned}
\lim\limits_{x \rightarrow 2} \quad \displaystyle \frac{2x^2+1}{x^2+6x-4} &= \displaystyle \frac{\lim\limits_{x \rightarrow 2} (2x^2+1)}
{\lim\limits_{x \rightarrow 2} (x^2+6x-4)}
&& \text{(Quotient Law)}\\
\lim\limits_{x \rightarrow 2} \quad \displaystyle \frac{2x^2+1}{x^2+6x-4} &= \displaystyle \frac{2\lim\limits_{x \rightarrow 2} x^2 + \lim\limits_{x \rightarrow 2} 1}
{\lim\limits_{x \rightarrow 2}x^2+6\lim\limits_{x \rightarrow 2}x-\lim\limits_{x \rightarrow 2}4}
&& \text{(Sum, Difference and Constant Law)}\\
\lim\limits_{x \rightarrow 2} \quad \displaystyle \frac{2x^2+1}{x^2+6x-4} &= \displaystyle \frac{2(2)^2+1}{(2)^2+6(2)-4}
&& \text{(Constant, Special Limit and Power Special Limit Law.)}
\end{aligned}
\end{equation}\\
\boxed{ \lim\limits_{x \rightarrow 2} \quad \displaystyle \frac{2x^2+1}{x^2+6x-4} = \displaystyle \frac{3}{4}}
$