Q.Evaluate $\displaystyle\lim_{x\to\infty}\frac{2x + 5}{3x^{2} - x + 1}$.

Numerator degree is smaller than denominator degree. …

Substitution gives $\tfrac{\infty}{\infty}$. Divide every term by $x^2$ (the denominator's highest power):

$\frac{\dfrac{2}{x}+\dfrac{5}{x^2}}{3-\dfrac{1}{x}+\dfrac{1}{x^2}}\xrightarrow{\ x\to\infty\ }\frac{0+0}{3-0+0}=\frac{0}{3}=0.$ …

When the top degree is lower than the bottom degree, the limit at infinity is $0$ — not the leading-coefficient rati …