This is what happens:
$\displaystyle \lim_{x \to \infty} \frac{5x+\sqrt{x^2+2x}}{4+x} = \lim_{x \to \infty} \frac{5x+\sqrt{x^2\left(1+\frac{2}{x}\right)}}{x \left(\frac{4}{x}+1\right)}$
$\displaystyle = \lim_{x \to \infty} \frac{5x+|x|\sqrt{1+\frac{2}{x}}}{x\left(\frac{4}{ x}+1\right)}$
Suppose $\displaystyle x \to +\infty$ then $\displaystyle |x| = x$ thus
$\displaystyle \lim_{x \to +\infty} \frac{x\left(5+\sqrt{1+\frac{2}{x}}\right)}{x\left (\frac{4}{x}+1\right)} = \lim_{x \to +\infty} \frac{5+\sqrt{1+\frac{2}{x}}}{\frac{4}{x}+1} = 6$
Suppose $\displaystyle x \to -\infty$ then $\displaystyle |x|=-x $
Now (check it) the limit will be equal to 4
Perphas this also can help you: substituting for example $\displaystyle x=-3$ in $\displaystyle \frac{5x+\sqrt{x^2+2x}}{4+x}$ we get $\displaystyle \frac{5(-3)+\sqrt{(-3)^2+2(-3)}}{4+(-3)}$. Now, divide numerator by $\displaystyle -\sqrt{(-3)^2}\;(=-3)$ and denonimator by $\displaystyle -3$ . What do you obtain? (substtitute now $\displaystyle -3$ by $\displaystyle x$)