# Asymptotic Analysis of $\log(1+x)$

I am trying to prove that $$\log(1+x) \leq \sqrt{x}$$ for all values $$x>x_0$$ which implies that $$\log(1+x) = O(\sqrt{x})$$. However, I am not able to get to this result. The result which states $$\log(1+x) = O(x)$$ is immediate as $$\log(1+x) \leq x, ~\forall x\geq0$$ which follows by expanding $$\log(1+x)$$ around $$0$$. However, I cannot show that $$\log(1+x)\leq \sqrt{x}$$ using the same approach.

Any help would be greatly appreciated!

For $$x>1$$ we have $$1/x < 1/\sqrt{x}$$. Integrating both sides will get you a relationship between the logarithm and $$\sqrt{x}$$ that can be massaged to get the desired result.

• Thanks for the comment. After your hint, I instantly derived it! Dec 27 '19 at 4:19

$$\ln (1+x)=\sqrt x$$ when $$x=0$$. Furthermore, when $$x\ge0$$, $$\dfrac d{dx} ln(1+x)\le \dfrac d{dx} \sqrt x$$ because $$\dfrac1{1+x} \le \dfrac1{2\sqrt x}$$ because $$2\sqrt x \le 1+x$$ because $$0\le(1-\sqrt x)^2$$.

Therefore, $$\ln(1+x)\le \sqrt x$$ when $$x\ge0$$.

• Thank you for your comment :) Dec 27 '19 at 4:24

The inequality you stated:

$$\ln(x)\le x-1\le x$$

Substitute $$x\mapsto\sqrt[4]x$$ and use log rules to get:

$$\ln(x)\le4\sqrt[4]x$$

But for $$x\ge4^4$$ we have

$$4\sqrt[4]x\le\sqrt x$$

• Thanks for your comment! This is also a very good way of showing it! Dec 27 '19 at 4:19