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.

  • 2
    $\begingroup$ What do you call "absolute zero" ?? $\endgroup$ – Yves Daoust May 18 '19 at 8:45
  • $\begingroup$ @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. $\endgroup$ – The_Sympathizer May 18 '19 at 11:20
  • $\begingroup$ Yes you are right @The_Sympathizer $\endgroup$ – The Learner May 18 '19 at 12:06

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.

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

  • $\begingroup$ 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😅 $\endgroup$ – The Learner May 18 '19 at 11:05
  • $\begingroup$ Thanks for the answer btw. $\endgroup$ – 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.

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

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