order of $ln(1+\epsilon)$ as $\epsilon \rightarrow 0$

Question

I am struggling with understanding how to find the order of this expression. The answer in the solutions is $\lim_{\epsilon \to 0}= \frac{ln(1+\epsilon)}{\epsilon}$. I don't understand the concept of putting $\epsilon$ in the denominator

Answer 1

Let us use this definition:

$$\log(1+\epsilon)=\int_1^{1+\epsilon}\dfrac{dt}t.$$

Then as the inverse is a monotonous function,

$$\frac{\epsilon}{1+\epsilon}=\int_1^{1+\epsilon}\dfrac{dt}{1+\epsilon}<\log(1+\epsilon)<\int_1^{1+\epsilon} dt=\epsilon.$$

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