Question

Find the value of $\underset{x\to \infty }{lim}\frac{8-\frac{3}{x}+\frac{5}{x^2}}{4+\frac{2}{x}+\frac{12}{x^2}}$

Answer 1

We have $\underset{x\to \infty }{lim}\frac{8-\frac{3}{x}+\frac{5}{x^2}}{4+\frac{2}{x}+\frac{12}{x^2}}$

We can directly substitute $x=\infty$ in the given expression

so, we have

$=\frac{8-\frac{3}{x}+\frac{5}{x^2}}{4+\frac{2}{x}+\frac{12}{x^2}}=\frac{8-0+0}{4+0+0}$

$=\frac{8}{4}=2$