Where is my mistake in evaluating $\lim_{x \to -\infty} \frac{\sqrt{x^2+4}}{x}$? Given is $\lim_{x \to -\infty} \dfrac{\sqrt{x^2+4}}{x}$
I divide numerator and denominator by x to the largest degree in the denominator and I get
$$\lim_{x \to -\infty} \frac{\sqrt{1+\frac{4}{x^2}}}{1}=\frac{\sqrt{1}}{1}=1.$$
But the answer should be -1. Where did I make a mistake?

The mistake is when you are shortening out $x$.  $x$ is negative, but the square root is positive.  Thus:
$$\begin{align}
\lim_{x \to-\infty} \frac{\sqrt{x^2+4}}{x}
&= \lim_{x \to-\infty} \frac{|x|\sqrt{1+4/x^2}}{-|x|} \\
&= \lim_{x \to-\infty} \frac{\sqrt{1+4/x^2}}{-1} \\
&\to -1
\end{align}$$
where $x=-|x|$ in the denominator because $x < 0$.

Alternative way to see the limit is substituting $x= -y$:
$$\begin{align}
\lim_{x \to-\infty} \frac{\sqrt{x^2+4}}{x}
&= \lim_{y \to\infty} \frac{\sqrt{(-y)^2+4}}{-y} \\
\end{align}$$
