Question

Using $L$ Hospital rule, evaluate ${lim}_{x\to \infty }\frac{e^x+{3x}^3}{{4e}^x+{2x}^2}.$

Answer 1

$\underset{x\to \infty }{lim}\frac{e^x+3x^3}{4e^x+2x^2}$

This is $\frac{\infty }{\infty }$ form, thus

applying L- Hospital Rule

$\therefore \underset{x\to \infty }{lim}\frac{e^x+9x^2}{4e^x+4x}$

again

$\Rightarrow \underset{x\to \infty }{lim}\frac{e^x+18x}{4e^x+4}$

again

$\Rightarrow \underset{x\to \infty }{lim}\frac{e^x+18}{4e^x+0}$

$\Rightarrow \underset{x\to \infty }{lim}\frac{1+18/e^x}{4+0}$ dividing $e^x$ both sides, we get

$\underset{x\to \infty }{lim}\frac{1+\frac{18}{\infty }}{4}\Rightarrow \frac{1+0}{4}\to \frac{1}{4}$