Limits at infinity of ratios of polynomials

As $x\to +\infty$ the numerator and the denominator both tend to $+\infty$, therefore the ratio is in the indeterminate form $\frac{\infty }{\infty }$. In each of them one factors out the highest-degree power, $x^2$, which then cancels since $x\ne 0$. The terms $\frac{1}{x}$, $\frac{5}{x^2}$ and $\frac{7}{x^2}$ tend to $0$, hence the ratio tends to $\frac{3}{2}$.

As $x\to +\infty$ the numerator and the denominator both tend to $+\infty$, therefore the ratio is in the indeterminate form $\frac{\infty }{\infty }$. One factors out the highest-degree power of each polynomial: $x^2$ in the numerator and $x^3$ in the denominator. After cancelling, what remains is the factor $\frac{1}{x}$, which tends to $0$, multiplied by a fraction that tends to $\frac{3}{1}=3$. Since the product of an infinitesimal function and a function with a finite limit is infinitesimal, the limit is $0$. The degree of the denominator exceeds that of the numerator, and this factoring shows why in that case the ratio always tends to $0$.

As $x\to -\infty$ the numerator tends to $+\infty$, since the power $x^4$ has an even exponent, whereas the denominator tends to $-\infty$, since $x^3$ has an odd exponent. One factors out the highest-degree power of each polynomial, $x^4$ above and $x^3$ below, and after cancelling what remains is the factor $x$ multiplied by a fraction. The terms $\frac{2}{x^3}$ and $\frac{1}{x^3}$ tend to $0$, therefore the fraction tends to $1$. The product of a function that tends to $-\infty$ and a function with limit $1$ tends to $-\infty$, hence the limit is $-\infty$. The sign of the result depends on the direction of infinity: as $x\to +\infty$ the same ratio instead tends to $+\infty$.