First slide

Introduction to limits

Question

$lim_{x \to \infty} \frac{x^{n}}{e^{x}} = 0, (n$ integer), for

Moderate

A

no value of $n$
B

all values of $n$
C

only negative values of $n$
D

only positive values of $n$

Solution

If $n$ is a negative integer, then $n = - m,$ where $m \in N .$
$∴ lim_{x \to \infty} \frac{x^{n}}{e^{x}} = lim_{x \to \infty} \frac{x^{- m}}{e^{x}} = lim_{x \to \infty} \frac{1}{x^{m} e^{x}} = 0$
If $n = 0,$ then
If $n \in N,$ then,
$lim_{x \to \infty} \frac{x^{n}}{e^{x}} = lim_{x \to \infty} \frac{n!}{e^{x}} = 0$ [By L' Hospital's Rule]
Hence, $lim_{x \to \infty} \frac{x^{n}}{e^{x}} = 0$ for all values of $n$ .

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