Question

evaluate the limit using lhopitals rule \\(\lim_{x\to\infty} \frac{8x^2}{e^{7x}}\\) question help: video message instructor submit question jump to answer

Explanation:

Step1: Check the form

When \(x\to+\infty\), \(\lim_{x\to+\infty}\frac{8x^{2}}{e^{7x}}\) is in the \(\frac{\infty}{\infty}\) form.

Step2: Apply L'Hopital's rule once

Differentiate the numerator and denominator. The derivative of \(y = 8x^{2}\) is \(y^\prime=16x\), and the derivative of \(y = e^{7x}\) is \(y^\prime = 7e^{7x}\). So \(\lim_{x\to+\infty}\frac{8x^{2}}{e^{7x}}=\lim_{x\to+\infty}\frac{16x}{7e^{7x}}\).

Step3: Apply L'Hopital's rule again

Differentiate the new - formed fraction. The derivative of \(y = 16x\) is \(y^\prime=16\), and the derivative of \(y = 7e^{7x}\) is \(y^\prime = 49e^{7x}\). So \(\lim_{x\to+\infty}\frac{16x}{7e^{7x}}=\lim_{x\to+\infty}\frac{16}{49e^{7x}}\).

Step4: Evaluate the limit

As \(x\to+\infty\), \(e^{7x}\to+\infty\). Then \(\lim_{x\to+\infty}\frac{16}{49e^{7x}} = 0\).

Answer:

\(0\)