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I know that $$\lim_{x\to 0} \dfrac{\ln(1+x)}{x}=1$$

I was trying to modify this result a bit.

Consider a function $f(x)$ such that $$\lim_{x\to a} f(x)=0$$

Substituting $x=f(x)$ in the original limit, $$\lim_{f(x)\to 0} \dfrac{\ln(1+f(x))}{f(x)}=1$$ $$$$ However I was wondering if I could rewrite $\lim_{f(x)\to 0}$ as $\lim_{x\to a}$ because as $x\to a$, $f(x)\to 0$ If this were correct, then I could have further modified $$\lim_{x\to 0} \dfrac{\ln(1+x)}{x}=\lim_{x\to a} \dfrac{\ln(1+f(x))}{f(x)}$$ $$$$ I'm not sure if replacing the condition $f(x)\to 0$ with $x\to a$ is allowed. I would be grateful if somebody would please explain why it is/is not allowed. Many thanks!

user342209
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  • What you have done is very natural way to evaluate limits and it is called the rule of substitution of limits. You only need to ensure that $f(x) \to 0$ but $f(x) \neq 0$ when $x \to a$. Thus if $f(x) = x\sin (1/x)$ and $a = 0$ then this is not valid because as $x \to 0$ you can see that $f(x) = 0$ at infinitely many points of the form $x = 1/n\pi$. If you have $f(x) = \sin x$ and $a = 0$ then it is perfect because $\sin x \neq 0$ as $x \to 0$. – Paramanand Singh May 25 '16 at 14:57
  • Sir from what I could understand you're saying that the condition for the substitution to be valid is that when $x\to a$ $f(x)\to 0$, BUT $f(a)\neq 0$. $$$$If we take $f(x)=\sin(x)$ and $a=0$ then as $x\to a$ doesn't $\sin(x)\to 0$? And isn't $\sin(0)=0$? In that case, how is $f(x)\to 0$ as $x\to a$ but $f(a)\neq 0$ ? – user342209 May 25 '16 at 15:48
  • Sir did you mean that $f(x)\neq 0$ in the neighbourhood of $x=a$ (excluding the point $x=a$)? – user342209 May 25 '16 at 15:55
  • I mean that $f(x)\neq 0$ in a certain neighborhood of $a$ except at $a$. So your last comment is correct. – Paramanand Singh May 25 '16 at 16:16
  • Alright Sir, thanks very much! – user342209 May 25 '16 at 16:52
  • @ParamanandSingh Sir, could you please answer my doubt here? It is quite similar to this one. – user342209 Jun 07 '16 at 17:51
  • Your other question already has good answers. The basic idea is to divide and multiply by $\sin u$ so that you have $\sin u$ in denominator. – Paramanand Singh Jun 08 '16 at 03:24

1 Answers1

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What you are doing is not always allowed. In fact, it is only allowed if $f(x)\neq 0$ on some set $(a-\delta, a+\delta)\setminus\{a\}$. I will, from now on, assume that $f$ satisfies that property.

You can see that what you are doing is allowed by going to the definitions. So, let's write down what you know:

you know that (property 1)

$$\lim_{x\to a} f(x)=0$$

which means that for every $\epsilon >0$, there exists such a $\delta >0$ that for all $x$ such that $0<|x-a|<\delta$, you have $|f(x)|<\epsilon$.

Also (property 2):

You know that $$\lim_{y\to 0}\frac{\ln(1+y)}{y}=1$$ which means that for every $\epsilon >0$, there exists such a $\delta>0$ that for all $x$ such that $0<|y|<\delta$, you have $\left|\frac{\ln(1+y)}{y}-1\right|<\epsilon.$

Now, you are trying to prove that $$\lim_{x\to a}\frac{\ln(1+f(x)}{f(x)}=1$$

Take the usual steps:

  1. Let $\epsilon > 0$ be arbitrary.
  2. Now, take $\delta_1$ to be the value $\delta_1$ for which, if $0<|x|<\delta_1$, you know that $\left|\frac{\ln(1+y)}{y}-1\right|<\epsilon.$
  3. Now, if you can only get $f(x)$ under $\delta_1$, you can finish your proof. But wait, you can! You can set $\delta_2$ such that if $0<|x-a|<\delta_2$, then $|f(x)|<\delta_1$ (because of property number 1, setting $\epsilon=\delta_1$ in that property. Therefore, you will now determine that your final $\delta$ is equal to $\delta_2$.

OK, so you found a $\delta$, but is it correct? Is it true that if $0<|x-a|<\delta$, then $\left|\frac{\ln(1+f(x))}{f(x)}-1\right|<\epsilon$?

Well, if $0<|x-a|<\delta=\delta_2$, then you know, by property 1 and the definition of $\delta_2$, that $|f(x)|<\delta_1$. Getting close?

5xum
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  • Thanks Sir. I'm trying to understand your solution. Sir, could you please explain what you meant by $$ (a-\delta, a+\delta)\setminus{a}?$$ I understood that $(a-\delta,a+\delta)$ represents an interval, but I couldn't understand the $ \setminus{a}$. Do you mean that we are to remove the element $a$ from the interval? – user342209 May 25 '16 at 14:03
  • Sir, I was able to understand whatever you wrote throughout the proof. Unfortunately I can't really understand the intuition behind the proof. I was able to understand that the various steps you undertook were correct, but I cannot understand why you undertook them. – user342209 May 25 '16 at 14:29
  • @user342209: The notation $A \setminus B$ denotes the set which contains all elements of $A$ except those which are elements of $B$ also. Thus in the current scenario the notation means that we have to remove the element $a$ from that interval and thus you are right about this. – Paramanand Singh May 25 '16 at 14:54
  • @user342209 Sorry for my late response, but I was away from my computer. The notation $(a-\delta, a+\delta)\setminus{a}$ means the set difference, so all numbers in the interval except $a$ – 5xum May 25 '16 at 17:42
  • @5xum Sir, could you please answer my doubt here? It is quite similar to the doubt you cleared here. – user342209 Jun 07 '16 at 17:52