Determine the limit of this function

[raopeng]

Homework Statement

find the limit of $x^x$when x approaches 0.

Homework Equations

The Attempt at a Solution

I am currently revising my Mathematical Analysis and this question pops up. The author only expects the reader to use the most fundamental knowledge of the limit, so I feel my method counts as "cheating"(rewriting as $e^{x lnx}, lim x lnx= 0$using the L'Hospital's Rule so the limit is 1). Is there an approach that involves "less complicated knowledge" because I really wish to fully grasp every notion of basic math? Thank you...

 

[HallsofIvy]

That's perfectly valid- you are using the the fact that exponential is a continuous function. That is, $\lim_{x\to 0} e^{f(x)}= e^{\lim_{x\to 0} f(x)}$. $x^x= e^{x ln(x)}$ and $x ln(x)$ goes to 0 as you say. The limit of $x^cx$ is $e^0= 1$.

 

[raopeng]

Thanks for the reply. That's my idea exactly. But that problem appears in a chapter well before the author introduces the L'Hospital's Theorem so there must be another way to do it... I am torturing myself...

 

[Mark44]

Thanks for the reply. That's my idea exactly. But that problem appears in a chapter well before the author introduces the L'Hospital's Theorem so there must be another way to do it... I am torturing myself...

I would check to see if the author has discussed this limit in the chapter you're in: $\lim_{x \to 0} x ln(x)$. Since he doesn't present L'Hopital's Rule until later, it's possible he has said something about this limit, which is the crux of the problem you're trying to solve.

 

[Infrared]

You don't really need l'Hopital for $\lim_{x\rightarrow 0} {xlnx}$. I would let $x = e^{-n}$ and take the limit as n→∞.
$\lim_{n\rightarrow \infty} {e^{-n}ln(e^{-n})} = \lim_{n\rightarrow \infty} {\frac {-n}{e^n}}$. I guess this limit is technically a l'Hopital problem, but it isn't hard to see that it goes to zero.

 

[raopeng]

Thank you, HS-Scientist, that's what I later did. It is a bit tedious because jumping from n to x also takes an extra proof so maybe in the end L'Hopital law is easier.

 

[Mark44]

You don't really need l'Hopital for $\lim_{x\rightarrow 0} {xlnx}$. I would let $x = e^{-n}$ and take the limit as n→∞.
$\lim_{n\rightarrow \infty} {e^{-n}ln(e^{-n})} = \lim_{n\rightarrow \infty} {\frac {-n}{e^n}}$. I guess this limit is technically a l'Hopital problem, but it isn't hard to see that it goes to zero.

It's not obvious to me that the latter limit is zero. This limit is an indeterminate form [-∞/∞], which means that L'Hopital's Rule applies, so how did you evaluate the limit without using L'H?

 

[Infrared]

It's not obvious to me that the latter limit is zero. This limit is an indeterminate form [-∞/∞], which means that L'Hopital's Rule applies, so how did you evaluate the limit without using L'H?

If your definition of e^x is its power series, then this is rather trivial. You can also use the idea that e^x is increasing faster than x without referencing l'Hopital. If we consider $\frac{x}{e^x}$ and replace x with x+r for any positive r>ln2 we get $\frac{x+r}{e^re^x}$. The numerator has increased by a factor of 1+r/x and the denominator by a factor of e^r, so we multiplied the fraction by $\frac{1+r/x}{e^r}$. Now for all x>(1+r)/2 (we can assume that this is the case because r is a constant and we can make x as large as we like), this ratio is smaller than 2/e^r. As long as we have r>ln2, this ratio is also less than one. So each time we increase x by r, we multiply it by something that is smaller than a constant which is less than one, so it must go to zero as we add r to the input an arbitrarily large number of times.

Edit: This argument makes an assumption which I forgot to state explicitly; namely that if the sequence $a_1,a_2,...a_n...$ goes to zero, then so does a function f such that $f(n)=a_n$ (In my argument, a_n is n/e^n evaluated at x,x+r...) This is not true in general, but here it works because of r being arbitrary (as long as it >ln2). This is because f(x) is always bounded from above by the smallest value of f more than ln2 less than x. So every time x is incremented by r, there is a new maximum value of f (smaller than the previous one) for all x greater than this new x.

 

[raopeng]

My method is similar to HS-Scientist. It first starts from the statement that $lim\frac{n}{a^n}=0$ where a > 0 and n is an integer, whose proof uses the fact that the sequence is bounded below and $lim \frac{a^{n+1}}{a^n}=\frac{1}{a}<1$. And then it passes to the continuous case so for any x we can take the integer part of it and construct two terms which sandwich x/e^x and have a common limit 0. Hence x/e^x approaches 0. But indeed it is too tedious..