lim _(xarrow infty )(x^5)/(e^x)

To evaluate the limit \(\lim_{x \to \infty} \frac{x^5}{e^x}\), we can use L'Hôpital's Rule, which is applicable for limits of the form \(\frac{\infty}{\infty}\).<br /><br />First, let's confirm that we have an \(\frac{\infty}{\infty}\) form:<br />\[<br />\lim_{x \to \infty} \frac{x^5}{e^x} = \frac{\infty}{\infty}<br />\]<br /><br />Since this is indeed an \(\frac{\infty}{\infty}\) form, we can apply L'Hôpital's Rule, which states that:<br />\[<br />\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}<br />\]<br />provided the limit on the right-hand side exists.<br /><br />Let \(f(x) = x^5\) and \(g(x) = e^x\). We need to find the derivatives of \(f(x)\) and \(g(x)\):<br />\[<br />f'(x) = 5x^4<br />\]<br />\[<br />g'(x) = e^x<br />\]<br /><br />Now, apply L'Hôpital's Rule:<br />\[<br />\lim_{x \to \infty} \frac{x^5}{e^x} = \lim_{x \to \infty} \frac{5x^4}{e^x}<br />\]<br /><br />We still have an \(\frac{\infty}{\infty}\) form, so we apply L'Hôpital's Rule again:<br />\[<br />\lim_{x \to \infty} \frac{5x^4}{e^x} = \lim_{x \to \infty} \frac{20x^3}{e^x}<br />\]<br /><br />Again, we have an \(\frac{\infty}{\infty}\) form, so we apply L'Hôpital's Rule once more:<br />\[<br />\lim_{x \to \infty} \frac{20x^3}{e^x} = \lim_{x \to \infty} \frac{60x^2}{e^x}<br />\]<br /><br />And once more:<br />\[<br />\lim_{x \to \infty} \frac{60x^2}{e^x} = \lim_{x \to \infty} \frac{120x}{e^x}<br />\]<br /><br />And one more time:<br />\[<br />\lim_{x \to \infty} \frac{120x}{e^x} = \lim_{x \to \infty} \frac{120}{e^x}<br />\]<br /><br />Finally, we have:<br />\[<br />\lim_{x \to \infty} \frac{120}{e^x} = \frac{120}{\infty} = 0<br />\]<br /><br />Thus, the limit is:<br />\[<br />\lim_{x \to \infty} \frac{x^5}{e^x} = 0<br />\]