Why is $ \lim_{x\to 0^+} \ln x = -\infty. $

Question

Simple question, can't seem to find an intuitive explanation anywhere. But the question is as follows, why does Why is $$ \lim_{x\to 0^+} \ln x = -\infty. $$

and Why is

$$ \lim_{x\to \infty} \ln x = \infty. $$ Honestly the second one is a tad bit easier to understand. You are trying to find the natural log of an infinite number so the number you would be left with, would also be infinite, but the first one ($\ln{0}$) just doest not make sense for me.

Thanks in advance.

Answer 1

I do not believe a statement like $\ln \infty = \infty$ has an exact, universally accepted mathematical meaning. What I think we can agree on is that whatever that formula represents, it is based on the truth of the generally-accepted true mathematical statement, $$ \lim_{x\to\infty} \ln x = \infty $$ which in turn is not really an equation, but is rather a way of saying that if we want to the value of $\ln x$ to be greater than some positive number (choose any number you want), all we have to do is to ensure that $x$ is a large enough positive number. There is no upper bound on how large we can force $\ln x$ to be, and all we have to do in order to make $\ln x$ "large enough" is name a number $N$ and assert that $x > N$.

So what we're really trying to explain is why $$ \lim_{x\to 0^+} \ln x = -\infty. $$ That is, to force $\ln x$ to be less than some arbitrarily large negative number, all we have to do is make $x$ close enough to (but greater than) $0$.

Now, we know that $$ \ln x = - \ln \frac1x, $$ and we know that $$ \lim_{x\to 0^+} \frac 1x = \infty. $$ So by making $x$ close to zero, but positive, we can make $\frac1x$ be as large a positive number as we like. And therefore we can make $\ln \frac1x$ be as large a positive number as we need to.

But if $\ln \frac1x$ is a large positive number, then $\ln x = -\ln \frac1x$ is a large negative number. And that's how we know that $$ \lim_{x\to 0^+} \ln x = -\infty. $$

Answer 2

Intuitively:

The function $y=\ln x$ is the inverse function of $y=e^x$, so its graph is obtained from the graph of $y=e^x$ interchanging the $x$ and $y$ axis ( that is: with a symmetry respect to the bisector of the first quadrant).

Answer 3

There's a mistake in your understanding. $$ln(\infty) \neq \infty$$ its actually undefined. The same applies to $$ln(0) \neq -\infty$$

Answer 4

just look at this: $ln(1/x) = -ln(x)$. So if $ x\to \infty $ , which means $ \frac{1}{x} \to 0$ we get $ln(1/x) \to - \infty$

Answer 5

So, what I understand the question to be is:

Why does $\lim_{x\to 0} \ln x=-\infty$ (since $\ln 0$ is undefined)

and

Why does $\lim_{x\to \infty} \ln x=\infty$ (since $\infty$ is not a number, thus cannot be evaluated at.)

Consider the function $f(x)=e^x$ as we allow $x$ to approach $-\infty$, that is get really really negative, then by using exponent properties we are dealing with a fraction $\frac{1}{e^{|x|}}$, so our function is getting really close to zero(but key that it doesn't equal zero.) Since $\ln x$ is the inverse of $e^x$ we have that $\lim_{x\to 0} \ln x=-\infty$

That same logic applies to $\lim_{x\to \infty} \ln x=\infty$. If we again consider $f(x)=e^x$ and we allow $x$ to apporach $\infty$ then $e^x$ gets really big, approaching infinity, and agina since $\ln x$ is the linverse of $e^x$ then we have that $\lim_{x\to \infty} \ln x=\infty$

Answer 6

The limit of the natural log as x approaches 0 is negative infinity. This is because you have to invert small positive numbers (less than 1) before exponentiating them if you want to send them to e ( a number greater than 1)

Answer 7

To be exact, $ln(0)$ does not equal $−∞$, but $\lim_{x\to0+}ln(x)=−∞$.
Think of it like this: $\log_a(x)=b$ is equivalent to saying $a^b=x$. Now you know that if $b$ approaches $−∞$ and $a > 1$, then $x$ approaches $0$ (look at the graph of an exponential function to verify). This is the case with $ln(x)$ because the natural logarithm has base $e$ and $e>1$.

Answer 8

The first step to this problem is to correct a phrasing. $\ln 0$ does not equal $-\infty$ and $\ln \infty$ does not equal $\infty$. 0 and infinity are not in the domain of ln. This is an important distinction to understand as one goes forward with calculus. The correct phrasing would be that $\ln x$ approaches $-\infty$ as x approaches 0 and $\ln x$ approaches $\infty$ as x approaches $\infty$. If we were to use the limit operator, which is the official way to talk about such things in calculus, we would say $lim_{x\to 0}\ln x = -\infty$ and $lim_{x\to \infty}\ln x = \infty$.

Because we're talking about limits, our intuitive argument can look at less extreme numbers and project towards the difficult cases of 0 and infinity (a rigorous approach would use delta-epsilon or some other approach, but you asked for intuition).

Let's start with the one that's more comfortable: $lim_{x\to \infty}\ln x = \infty$. Let's look at $log x = y$ for varying values of x. I've switched to base 10 only because it makes it easier to write clean examples, but the same arguments work for natural logarithms. log 10 = 1, log 100 = 2, log 1000 = 3, log 10000 = 4. We can see that base-10 logarithms are roughly counting the number of digits in a number (very roughly, but its' good enough for our intuition here). Thus, if we were to ask what $\ln x$ equals as x approaches infinity, a reasonable answer would be "roughly equal to the number of digits in infinity." Of course, the concept of "the number of digits in infinity" is as meaningless as $\ln \infty$ was, but it points intuitively towards the right answer. As x grows without bound, so does the number of digits in x. Intuitively, this should be close enough to $lim_{x\to \infty}\ln x = \infty$ to make a leap. To be more exacting, seek a formal answer.

Likewise, we can make a similarly intuitive argument for $lim_{x\to 0}\ln x = -\infty$. Consider log 1 = 0, log 0.1 = -1, log 0.01 = -2, log 0.001 = -3, and so forth. In a very informal way, we can say that this base 10 logarithm is roughly the number of digits after the decimal place of the first non-0 digit, times -1 to make it a negative number. So as you get closer and closer to 0, say 0.000000001 or 0.000000000000000001, you will keep counting more and more digits. As you approach 0, the number of 0's in the front of the number grow without bound (just like how the number of digits in the case of logarithms of large numbers grew without bound).

There are certainly formal proofs which capture this with substantially more rigor, but intuitively, these arguments about values "growing without bound" should help!

Answer 9

Starting with $$\lim_{x\to - \infty}e^x=$$ $$e^{- \infty}=\frac {1}{e^ \infty}=\frac {1}{\infty}=0 $$ Apply $\rm ln$ into both sides of this equality. $$\ln({e^{- \infty}})=\ln0$$ Then $$- \infty=\ln0$$ For the second, just use the same reasoning, now with positive $\infty$.