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Question

I wanted to prove

lim_{x \to \infty} \frac{\ln x}{x} = 0

by the squeeze theorem.

Answer 1

\ln x = 2 \ln x^{\frac{1}{2}} \leq 2 \sqrt{x}

(using the inequality

y \geq \ln y

for

y > 0

that you obtained.)
So it follows that: for all

x > 0

| \frac{\ln x}{x} | \leq 2 \frac{\sqrt{x}}{x}

It follows by squeeze theorem that

lim_{x \to \infty} | \frac{\ln x}{x} | = 0.

Now note that

- | \frac{\ln x}{x} | \leq \frac{\ln x}{x} \leq | \frac{\ln x}{x} |

and hence the result follows by squeeze theorem.

Answer 2

Extending the approximation

e^{x} \geq 1 + x

into

e^{x} \geq 1 + x + \frac{x^{2}}{2} > \frac{(x + 1)^{2}}{2}

and

e^{x} \geq 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} > \frac{(x + 1)^{3}}{6}

allows

e^{x} > \frac{(x + 1)^{n}}{n!}

for all

n \in Z

So,

x \geq n \ln (1 + x) - \ln n! \geq n \ln x - \ln n!

So

\frac{x}{n} + \frac{\ln n!}{n} \geq \ln x

\frac{1}{n} + \frac{\ln n!}{n x} \geq \frac{\ln x}{x}

As we can make both terms on the LHS arbitrarily small, we are done.

Answers (2)