Solution.

Direct substitution yields \(\frac {\infty }{\infty }\). We therefore divide both the numerator and denominator by the highest power of \(x\).

\begin {align*} \lim _{x\longrightarrow \infty } \frac {2x}{1+3x} & = \lim _{x\longrightarrow \infty } \left (\frac {2x/x}{1/x+3x/x}\right )\\\\ & = \lim _{x\longrightarrow \infty } \frac {2}{1/x+3}\\ & = \frac {\lim \limits _{x\longrightarrow \infty }2}{\lim \limits _{x\longrightarrow \infty }1/x+\lim \limits _{x\longrightarrow \infty }3}\\ &=\frac {2}{0+3}=\frac {2}{3}\\ \end {align*}

Solution. Divide above and below by \(x^{2}\), the highest power present, in both cases. \begin {align*} \lim _{x\to \infty }\frac {5-2x+9x^{2}}{2x^{2}+3} &=\lim _{x\to \infty }\frac {\frac {5}{x^{2}}-\frac {2}{x}+9}{2+\frac {3}{x^{2}}} =\frac {0-0+9}{2+0}=\frac {9}{2},\\[6pt] \lim _{x\to \infty }\frac {1+5x-3x^{2}}{7x^{2}+2x} &=\lim _{x\to \infty }\frac {\frac {1}{x^{2}}+\frac {5}{x}-3}{7+\frac {2}{x}} =\frac {0+0-3}{7+0}=-\frac {3}{7}. \end {align*}