Limits at infinity of rational and irrational functions

The numerator tends to $+\infty$ and the denominator to $-\infty$, so the form is $\frac{\infty }{\infty }$. In both we factor out the highest power, $x^3$, and cancel it. The terms $\frac{1}{x^2}$ and $\frac{1}{x^3}$ tend to $0$, therefore the quotient tends to the ratio of the coefficients of $x^3$, that is $-\frac{3}{2}$. The negative sign is consistent with the fact that, for large $x$, the numerator is positive and the denominator negative.

We factor out the highest powers as for $x\to +\infty$, paying attention to the signs. In the first quotient the factor $x$ remains, which tends to $-\infty$, multiplied by a factor that tends to $1$; the limit is therefore $-\infty$. In the second, which is the reciprocal of the first, the factor $\frac{1}{x}$ remains, which tends to $0$, and the limit is $0$. The result is consistent with the sign, because for very negative $x$ the numerator $x^2+5$ is positive and the denominator $x+1$ is negative, so the first quotient is negative, like its limit.

The root and $x$ both tend to $+\infty$, and the difference is in the form $+\infty -\infty$. We rationalize by multiplying and dividing by $\sqrt{x^2+4x}+x$, and so the numerator becomes $4x$. In the denominator we bring the factor $x^2$ out of the root, which comes out as $x$ because $x>0$. After cancelling, the root tends to $1$ and the limit is $2$. The result means that, for large $x$, $\sqrt{x^2+4x}$ is approximately $x+2$.