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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

user643396
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  • Fix the typo in your equation, this is meaningless. –  Apr 18 '19 at 12:20
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    Since $\ln(1) = 0$, this is the same as $\lim_{\epsilon \to 0} \frac{\ln(1+\epsilon) - \ln(1)}{\epsilon}$. Does this remind you of something? – NickD Apr 18 '19 at 12:23
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    Usually the order of $f(x)$ as $x \to 0$ means the largest real number $a$ such that $\frac{f(x)}{x^a}$ remains bounded as $x \to 0$. That's more or less just the definition, though if you have an actual definition in hand then that's helpful to us for explaining things. In any case, do you see why it makes some sense to guess that the order might be $1$ here? – Ian Apr 18 '19 at 12:23
  • I believe I have to use the Gauge function which is defined as $g(\epsilon)$ and is used in the context $\lim_{\epsilon \to 0} \frac{f(\epsilon)}{g(\epsilon)}=A$ where A is a non-zero constant. – user643396 Apr 18 '19 at 12:34

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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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