# Conceptual doubt on “Limits”

Suppose I have a function say $$f(x)= x^2$$ . Now we know that graph is parabola, and it passes through the origin.

Now I write $$x^2$$ as $$e^{2 ln(x)}$$ . I plug in the value $$0$$ . I know that $$ln 0$$ approaches $$-\infty$$ . So my answer should be $$\frac{1}{e^{2\infty}}$$ which is approaching zero . However in the situation above, I am getting zero.

It seems, I am having some problem in understanding the concept. Any help would be greatly appreciated. Approaching zero and zero, I guess are not same.

• What do you call "absolute zero" ?? – Yves Daoust May 18 '19 at 8:45
• @Yves Daoust: I think he means "equalling zero". Having a limit of zero and having a value of zero at a point are two different things. – The_Sympathizer May 18 '19 at 11:20
• Yes you are right @The_Sympathizer – The Learner May 18 '19 at 12:06

## 3 Answers

First of all, if absolute $$0$$ is the absolute value of $$0$$, then note that $$\lvert0\rvert=0$$. So, there is no difference between zero and absolute zero.

On the other and, if you write $$x^2$$ as $$e^{2\ln x}$$, then you have a problem: since$$\ln x$$ doesn't exist when $$x<0$$, from the fact that $$\lim_{x\to0}e^{2\ln x}=0$$ all you can deduce is that $$\lim_{x\to0^+}x^2=0$$. Of course, since $$x^2$$ is an even function, it follows from this that $$\lim_{x\to0^-}x^2=0$$ too.

• I think what OP means is "approaching zero" is $\lim_{x\to c} f(x) = 0$ and "absolute zero" is $f(c) = 0$. – Infiaria May 18 '19 at 8:35
• Oh yes @José Carlos Santos , perfect explanation. I actually wanted to show how that left hand limit is zero. Thanks man. – The Learner May 18 '19 at 8:42
• I'm glad I could help. – José Carlos Santos May 18 '19 at 8:43

$$x^2,e^{2\ln(x)}\text{ and }e^{\ln(x^2)}$$ are three different things.

The first is always defined, the second requires $$x>0$$ and the third $$x\ne0$$. But all three have the limit $$0$$, because only the points inside the domain matter.

As far as I know, there is nothing commonly defined as "absolute zero".

• I think I should reword my question , I actually mean to say that absolute zero is zero which is different than approaching zero, I actually wanted to think this question in terms of limits @Yves Daoust, that's why I have written absolute zero . Nothing more than this😅 – The Learner May 18 '19 at 11:05
• Thanks for the answer btw. – The Learner May 18 '19 at 11:08

You consider the functions $$f : \mathbb R \to \mathbb R, f(x) = x^2$$, and $$g : (0,\infty) \to \mathbb R, g(x) = e^{2\ln x}$$. Note that $$\ln x$$ does not exist for $$x \le 0$$. In particular, $$\ln 0$$ does not exist. You write "$$\ln 0$$ approaches $$-\infty$$", but in this form it does not make sense. What you can say is that $$\lim_{x \to 0+} \ln x = -\infty$$ and therefore $$\lim_{x \to 0+} g(x) = 0$$.

We have $$f(x) = g(x)$$ for $$x > 0$$. In this situation, without looking at the concrete definitions of $$f, g$$, we can be sure that $$\lim_{x \to 0+} f(x)$$ exists if and only $$\lim_{x \to 0+} g(x)$$ exists, and if these limits exist, they are equal. In the concrete case $$\lim_{x \to 0+} x^2 = 0$$, hence also $$\lim_{x \to 0+} e^{2\ln x} = 0$$. For this conclusion we do not need to know that $$\lim_{x \to 0+} \ln x = -\infty$$ and $$\lim_{y \to -\infty} e^{2y} = 0$$.

I guess that your wording "absolute zero" means $$f(0) = 0$$ in contrast to "approaching zero" which means $$\lim_{x \to 0+} g(x) = 0$$. But be aware that this causes confusion, in particular "absolute zero" is really opaque.

• Your analysis of absolute zero is correct @Paul Frost. Btw thanks for the answer. – The Learner May 18 '19 at 11:07