-1

Why the value of $\lim_{x\to 0} (1+x)^{1/x}$ is $e$; where $e$ is Euler's number($\approx 2.71$), whereas on the other hand $1^{\lim_{x\to \infty} x}$ is undefined. I can't understand that why the value of the 1st limit exists whereas on the other hand the value of the 2nd limit doesn't exists. Also both the limits are of the form of $1^{\infty}$ then why this discrepancy is occurring?

Shivam Kumar
  • 381
  • 1
    You have severe formatting issues and missing information for your second limit. As it reads now it is $1$ to the power of a limit of an empty expression. Assuming you were to change this to $\lim\limits_{x\to \infty} 1^x$ or to $1^{\lim\limits_{x\to \infty} x}$ or similar, then these are not undefined and the limit does exist... those are both identically equal to $1$. Yes, they are both "of the form $1^\infty$ but the punchline is that in the one case it is not exactly 1 raised to a large number while in the second case it is exactly one raised to a large number, very different. – JMoravitz Oct 14 '23 at 15:31
  • That is, assuming you allow for the second expression I wrote in the first place and/or are working in the extended real numbers, allowing for infinity as a number. – JMoravitz Oct 14 '23 at 15:32
  • @JMoravitz if you check on Wolfram Alpha then 1^{\lim: \lim_{x\to \infty} x} is undefined. It is not defined. – Shivam Kumar Oct 14 '23 at 15:50
  • 2
    You put too much faith in calculators and need to think for yourself. You need to understand that multiple conventions and possible definitions/interpretations can exist. I already qualified my statement that the second expression is poorly written, and if allowed requires certain interpretations to be valid. – JMoravitz Oct 14 '23 at 15:52
  • Ok. Yes you are correct@JMoravitz. – Shivam Kumar Oct 14 '23 at 15:53
  • The root of your confusion is probably related to what indeterminate forms are in the first place and in correctly interpreting what people mean when they write "$\infty \times 0$" or "$1^\infty$" and the like to really be things like $\lim\limits_{x\to \infty}\left(f(x)\times g(x)\right)$ when $f(x)\to \infty$ and $g(x)\to 0$ and the like, and comparing this to the case of the individual parts being "moving parts" versus stationary constants. Again, refer to my earlier link as well as https://math.stackexchange.com/questions/28940/why-is-infty-cdot-0-not-clearly-equal-to-0 and others here – JMoravitz Oct 14 '23 at 16:00
  • 1
    With regards to why wolfram might have said that $1^{\lim\limits_{x\to\infty}x}$ might be undefined... that could be because it chose to interpret it as though we were working in the real numbers and not in the extended real numbers... meaning that $\infty$ is not a number and so $1^\infty$ should make as much sense as $1^{\heartsuit\text{jabberwocky}\heartsuit}$. Raising $1$ to the power of something which is not a number shouldn't be allowed. Of course, with different contexts and interpretations, $1^\infty$ can be perfectly allowed, especially with $1$ as literal exact unchanging $1$. – JMoravitz Oct 14 '23 at 16:03

1 Answers1

0

P.S: A mistake down there, 'x' to the power infinite, when that 'x' lies between 0 and 1 is 0 as well.

I think you have a misconception here, first let me clear that. The answer to your second question is not undefined, it's 1. Anything to the power 1 is 1 even if it's infinite.

ONLY when the base is APPROACHING (not exact, it should be approaching from RHS or LHS) 1 while the exponent is approaching infinity makes 1 to the power infinity an indeterminant form. And other numbers except 1 and 0 to the power infinity is undefined/ infinity.

Now coming to your question, the first limit is variable dependent on both base and exponent, but for second limit only the exponent is variable dependent, this is the main difference between the two. Any CONSTANT (except 0 and 1) to the power infinity will be infinity/ undefined, I think that's intuitive.

Before talking about the first limit again let us get 2 facts clear, 1 to power infinity is 1 and any other number to the power infinity (0 and 1 excluded) is infinity.

So in first limit if you look closely, the base is approaching 1 while the exponent is approaching infinity. Doesn't it sound eerily similar to limit of 1 to the power of infinity which is taught to be indeterminant form? And what do we do with IF? We manipulate it to get a finite answer, if possible.

So in case of $lim x→0 (1+x)^{1/x}$

We do, $\lim: \lim_{x\to 0} e^{ln(1+x)^{1/x}}$

$\lim: \lim_{x\to 0} e^{ln(1+x)/x}$

$\lim: \lim_{x\to 0} e^{1}$

=e

I think this should clear your doubts regarding this. But I believe there are much more rigorous approach to explain your query but I am not knowledgeable enough for that.

Rhythmic_Ocean
  • 61