Question

Evaluate ${lim}_{x\to \infty }\frac{2x^2-3x+5}{7x^3-8x-4}$

Answer 1

We need to find value of limit ${lim}_{x\to \infty }\frac{2x^2-3x-5}{7x^3-8x-4}$

$={lim}_{x\to \infty }\frac{2-\frac{3}{x}-\frac{5}{x^2}}{7x-\frac{8}{x^2}-\frac{4}{x^3}}=x\to \infty \Rightarrow \frac{1}{x}\to 0,\frac{1}{x^2}\to 0,\frac{1}{x^3}\to 0={lim}_{x\to \infty }\frac{2-0-0}{7x-0-0}=\frac{2}{7}{lim}_{x\to \infty }\frac{1}{x}=\frac{2}{7}\times 0=0$