Again we have a situation where the numerator and denominator both approach \(\infty\text{.}\) Using the same intuition as above, we guess that the \(x^2\) term under the radical is the dominant one, as is the \(x^2\) term in the denominator. We make rigorous this intuition by factoring these out. Note how we must be careful with the radical simplification. In particular, we use the fact that \(\sqrt{x^2}=\abs{x}\text{,}\) and then deal with the absolute value appropriately:
\begin{align*}
\lim\limits_{x\to -\infty}\frac{\sqrt{x^2+4}}{2x^2+1} \amp = \lim\limits_{x\to -\infty}\frac{\sqrt{x^2(1+4/x^2)}}{x^2(2+1/x^2)} \\
\amp = \lim\limits_{x\to -\infty}\frac{\abs{x}\sqrt{1+4/x^2}}{x^2(2+1/x^2)}\\
\amp =\lim\limits_{x\to -\infty}\frac{-x\sqrt{1+4/x^2}}{x^2(2+1/x^2)} \amp (x\leq 0\implies \abs{x}=-x)\\
\amp = \lim\limits_{x\to -\infty }-\frac{1}{x}\cdot\frac{\sqrt{1+4/x^2}}{x(2+1/x^2)}\\
\amp = -\lim\limits_{x\to -\infty }\frac{1}{x}\cdot \frac{-\sqrt{1+4 \lim\limits_{x\to -\infty}1/x^2}}{2+\lim\limits_{x\to -\infty}1/x^2} \amp \text{(prod., quot., root, sum)}\\
\amp = 0\cdot \frac{\sqrt{1+0}}{2+0} \amp (1/x^n\to 0)\\
\amp = 0\text{.}
\end{align*}