*Using epsilon definition of the limit Prove that limn!(3) =n0

Answered: Prove that lim n! n n = 0

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

*Using epsilon definition of the limit

Expert Solution

Step 1: Proof

$\underset{}{\lim f (n)} = \lim_{} (\frac{n!}{n^{n}})$

We know that $h (\lim_{} g (x)) = \lim_{} f (g (x))$

Therefore

$\log (\lim_{} f (n)) = \lim_{} \log (\frac{n!}{n^{n}}) = \lim_{} \log (\frac{1}{n}) + \lim_{} \log (\frac{2}{n}) + . . . . . + \lim_{} \log (\frac{n}{n})$

Now we have to show that for $1 \leq m \leq n$ , $\lim_{} \log (\frac{m}{n}) = 0$ holds which is enough to say that $\underset{}{\lim f (n)} = \lim_{} (\frac{n!}{n^{n}}) = 0$

From the definition of of limit that a function $f$ is said to be tends to $l$ as $x \to \infty$ if for each $ε > 0$ there exist $k > 0$ such that $|f (x) - l| < ε,$ when $x > k$

Let choose a positive $ε$

Now $|\log (\frac{m}{n}) - 0| < ε$ will hold if $\frac{m}{n} < e^{ε},$ that is $n > \frac{m}{e^{ε}}$ .

Now let $k = \frac{m}{e^{ε}}$ and we know $1 \leq m \leq n$ and $e^{ε} > 1$ , clearly $\frac{m}{e^{k}} < n$ , therefore $n > k$

Therefore for $|\log (\frac{m}{n}) - 0| < ε$ there exist $n \geq k$

Hence the limit exist and the function $f (n) \to 0$ as $n \to \infty$ if for each $ε > 0$ there exist $k > 0$ such that $|f (n) - 0| < ε,$ when $n > k$