mathop {lim }limits_{x to infty } frac{{{x^n} | vedclass

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(B) Using $L$'$H$ôpital's Rule repeatedly,we differentiate the numerator and denominator $n$ times:$\mathop {\lim }\limits_{x 	o \infty } \frac{{{x^n

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$n$ is any whole number

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(B) Using $L$'$H$ôpital's Rule repeatedly,we differentiate the numerator and denominator $n$ times:$\mathop {\lim }\limits_{x 	o \infty } \frac{{{x^n}}}{{{e^x}}} = \mathop {\lim }\limits_{x 	o \infty } \frac{{n \cdot x^{n-1}}}{{{e^x}}} = \mathop {\lim }\limits_{x 	o \infty } \frac{{n(n-1) \cdot x^{n-2}}}{{{e^x}}} = \dots = \mathop {\lim }\limits_{x 	o \infty } \frac{{n!}}{{{e^x}}}$.Since $\mathop {\lim }\limits_{x 	o \infty } e^x = \infty$,we have $\frac{n!}{\infty} = 0$.This holds true for any whole number $n \ge 0$.

$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{e^x}}} = 0$ for

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