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 $x= -y$:
$$\begin{align} \lim_{x \to-\infty} \frac{\sqrt{x^2+4}}{x} &= \lim_{y \to\infty} \frac{\sqrt{(-y)^2+4}}{-y} \\ \end{align}$$