evaluate the limit using lhopitals rule. lim(x→+∞) 7x² / e³x =

Answer

Explanation:

Step1: Check indeterminate - form

As (x\to+\infty), (\lim_{x\to+\infty}\frac{7x^{2}}{e^{3x}}) is in the (\frac{\infty}{\infty}) form since (\lim_{x\to+\infty}7x^{2}=\infty) and (\lim_{x\to+\infty}e^{3x}=\infty).

Step2: Apply L'Hopital's rule

Differentiate the numerator and denominator. The derivative of (y = 7x^{2}) is (y^\prime=14x) and the derivative of (y = e^{3x}) is (y^\prime = 3e^{3x}). So, (\lim_{x\to+\infty}\frac{7x^{2}}{e^{3x}}=\lim_{x\to+\infty}\frac{14x}{3e^{3x}}). This is still in the (\frac{\infty}{\infty}) form.

Step3: Apply L'Hopital's rule again

Differentiate the new - numerator and denominator. The derivative of (y = 14x) is (y^\prime=14) and the derivative of (y = 3e^{3x}) is (y^\prime = 9e^{3x}). So, (\lim_{x\to+\infty}\frac{14x}{3e^{3x}}=\lim_{x\to+\infty}\frac{14}{9e^{3x}}).

Step4: Evaluate the limit

As (x\to+\infty), (e^{3x}\to+\infty), so (\lim_{x\to+\infty}\frac{14}{9e^{3x}} = 0).