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).


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