# Funky limit question

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Today's calculus warmup question has stumped everyone in the class, including our teacher. Here's the question:

$\lim_{x\to 0} x^{\sin x}$

We can't think of any safe way to evaluate the limit. We tried rewriting the problem to this:

$y = \lim_{x\to 0} x^{\sin x}$

and then changing to this:

$\ln y = \lim_{x\to 0}\sin x \ln x$

but in the end it didn't help.

Any ideas?

[hide]The answer turns out to be 1, but we have no idea to get it.[/hide]

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This is probably wrong, but at first glance, as x goes to 0 then sinx = sin0 = 0.

so x^0 = 1. (as far as Im aware - even 0^0 = 1 no?)

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I'd be very tempted to take a taylor series expansion of the function...

Not sure if that'd actually help though

00 is I suspect undefined, but I honestly don't know, the general thought of the 5 others in the room is that it's 1, but none of us know or are mathematicians

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I think the answer is indeed $1$.

DrP and Klay are almost right. We are looking at

$\lim_{x \rightarrow 0} x^{x}$,

which is $1$.

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Yeah, $0^0$ is indeterminate. I guess you can take the limit though, as ajb points out. It's deceptively simple.

Thanks, everyone.

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Yes, the limit is very important.

Try this for size,

$\lim_{x\rightarrow 0} x^{\cos(x)} = ?$

Again just series expand cos near zero and your away.

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So essentially what's happening is we decide (as in your example) that $\lim_{x \to 1} \cos x = 1$ and thus we can substitute in 1 in the original limit?

Or is there some limit technique I don't yet know?

Edited by Cap'n Refsmmat
multiple post merged

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All this works out because cos and sin have nice series expansions about zero. So, basically what you say is correct. However, I don't think it is always that simple.

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The trouble is, we haven't learned about series expansions yet. So is there an alternative method or was this question really just shown to us too soon?

What we've been learning, and the section of the textbook this problem was from, is L'Hopital's Rule. Our approach was to try to get the problem into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and work from there, but there's no clear way to do that with this problem. If it could be done, it'd make a lot more sense.

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Yeah, $0^0$ is indeterminate.

It's "less indeterminate" than $\frac{0}{0}$ though.

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and then changing to this:

$\ln y = \lim_{x\to 0}\sin x \ln x$

but in the end it didn't help.

You just want to rewrite it as $\frac{\infty}{\infty}$ by doing

$\ln y = \lim_{x\to 0}\frac{\ln x}{\frac{1}{\sin x }}$

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No, that's not $\frac{\infty}{\infty}$ because $\lim_{x \to 0} \frac{1}{\sin x}$ doesn't exist.

It turns out the natural log method would have worked if we were cleverer:

$\ln y = \lim_{x\to 0}\sin x \ln x$

$\ln y = \lim_{x\to 0} \frac{\sin x}{(\ln x)^{-1}}$

apply L'Hopital's Rule, since the above limit is indeterminate:

$\ln y = \lim_{x\to 0} \frac{\cos x}{-(\ln x)^{-2} \times x^{-1}}$

Simplificate:

$\ln y = -1 \times \lim_{x\to 0} \cos x \times \lim_{x\to 0} (\ln x)^2 \times x$

$\ln y = -1 \times \lim_{x\to 0} (\ln x)^2 \times x$

$\ln y = -1 \times \lim_{x\to 0} \frac{(\ln x)^2}{x^{-1}}$

L'Hopital again:

$\ln y = -1 \times \lim_{x\to 0} \frac{2\ln x \times x^{-1}}{-1x^{-2}}$

$\ln y = -1 \times \lim_{x \to 0} \frac{2\ln x}{-x^{-1}}$

...and again:

$\ln y = -1 \times \lim_{x \to 0} \frac{2x^{-1}}{x^{-2}}$

$\ln y = -1 \times \lim_{x \to 0} 2x$

$\ln y = \lim_{x \to 0} -2x$

$\ln y = 0$

$y = 1$

If anyone sees an error, please tell. I'd like to think I've conquered this one (with a bit of help).

And a question: I find it hard to believe $\lim_{x \to 0} \frac{\ln x}{x^{-1}} = \frac{\infty}{\infty}$. The bottom doesn't have a definite limit (I think) as you approach zero, so can you safely call it $\infty$ anyway to make the limit indeterminate?

Edited by Cap'n Refsmmat

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I think that infinity counts as infinity, even if you don't know whether it is positive or negative. The plan is to get rid of the infinity in the first place. Or you can do like you did, just invert it to get 0/0.

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$\lim_{x\to 0} x^{\sin x}$

$0^{0}$ is indeed indeterminate, so yeah, rewrite it.

$\lim_{x \to 0}\;e^{\displaystyle \ln x^{\sin x}}$

You use the ln to bring the sinx out in front, but you need the e to balance that. A lot of people just take the natural log of both sides, which works too. In my notes/homework I like working across the page instead of having a new line, and this reduces clutter.

$e^{\displaystyle \lim_{x\to0}\;\sin x \ln x}$

In this step, I bring the sinx out before the ln, and the limit inside the exponent. Notice you have $0\times-\infty$

$e^{\displaystyle \lim_{x\to0}\;\frac{\ln x}{\frac{1}{\sin x}}}$

We're dealing with the $\frac{\infty}{\infty}$ form, so we use l'Hopital's rule.

$e^{\displaystyle \lim_{x\to0}\;\frac{\frac{1}{x}}{\frac{-\cos x}{sin^{2}x}}}$

Simplified...

$e^{\displaystyle \lim_{x\to0}\;\frac{sin^{2}x}{-x\cos x}}$

Now we have $\frac{0}{0}$ so we use l'Hopital's rule again.

$e^{\displaystyle \lim_{x\to0}\;\frac{2\sin x \cos x}{x\sin x - \cos x}}$

The limit approaches $\frac{0}{-1}$, or just 0

$e^{0}\;\;=\;\; 1$

It's also important to note that for the function $f(x)=x^{\sin x}$

the Domain is $(0,\infty)$, so technically $\lim_{x\to0}\;x^{\sin x}$ doesn't exist, only $\lim_{x\to0^{+}}$ does.

$-1 \times \lim_{x\to 0} \cos x \times \lim_{x\to 0} (\ln x)^2 \times x \neq \; -1 \times \lim_{x\to 0} (\ln x)^2 \times x$

You cannot take the limit of part of the equation, you must wait until the end to take the entire limit at one time.

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Why is that? From what I understand of limits, the following is true:

$\lim_{x\to a} (f(x) \times g(x)) = \lim_{x\to a}( f(x)) \times \lim_{x\to a} (g(x))$

and you can evaluate each part safely.

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Yeah, you can distribute the limit, but you can't take them individually.

For the mathematical reason, I'm not entirely sure of the details, other than my Calc 1 professor used to say "I could prove a lot of untrue things if I did that".

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That seems contradictory. If $\lim_{x\to a} (f(x) \times g(x)) = \lim_{x\to a}( f(x)) \times \lim_{x\to a} (g(x))$, it's pretty much implied that you can take the limits individually -- why else would you have split the limit in two?

I'd like to see the explanation for this. I guess there are lots about limits I have yet to know...

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Okay, I'm seeing it now.

The limit laws only work when $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist.

In this case, $\lim_{x\to 0}\; x\ln^{2} x$ doesn't exist, so you can't use that limit law.

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So if we look at a point on the "end" of the domain of the function -- such as 0 in our example -- we cannot take the limit because we do not have two "sides" to approach from?

My understanding was that you just took the limit from the "side" you have and called it good. I will have to dig out my textbook tomorrow and see what else I'm mixing up with limits.

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Say we have $\lim_{x\to a} n$ to make this incredibly vague;

If $\lim_{x\to a^{+}} n \neq \lim_{x\to a^{-}} n$ then we say the limit doesn't exist at n. Which makes sense, since either side of the graph is approaching a different value. If you're trying to find the right/left-hand limit, then that's fine. But if it's not specified, it's assumed to be from both sides.

For this problem, I'd say that $\lim_{x\to 0^{+}}\; x^{\sin x} \; = \; 1$

The domain has nothing to do with the limit laws, if that was implied. I'm guessing the problem was looking for the answer of 1 regardless, and the right-hand limit wasn't important in evaluation.

Edited by darknecross
typo

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Just use l' opitals rule!

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For this problem, I'd say that $\lim_{x\to 0^{+}}\; x^{\sin x} \; = \; 1$

The domain has nothing to do with the limit laws, if that was implied. I'm guessing the problem was looking for the answer of 1 regardless, and the right-hand limit wasn't important in evaluation.

I'm a bit late to the discussion, but I'd like to point out that this is entirely correct. When taking a limit, you're considering regions centered around the point of interest, so your function at the very least must be defined there. Now, in your proof, you take [imath]\lim_{x\to 0} \log x[/imath] which is not defined, since log is only defined on the half-line [imath](0,\infty)[/imath]. So the evaluation of that limit is not possible.

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% Required matlab code
x = -20:0.01:20;
f = x.^(sin(x));
plot(x,f);grid on;


Hi guys , i was wondering what the result is and decided to code it in matlab to see correct result and it is 1. =()

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Today's calculus warmup question has stumped everyone in the class, including our teacher. Here's the question:

$\lim_{x\to 0} x^{\sin x}$

${{x}^{\sin x}}=\exp \left( \sin (x)\ln x \right)=\exp \left( \frac{\sin x}{x}\cdot x\ln x \right)\to 1$ as $x\to0,$ since

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1$ and $\underset{x\to 0}{\mathop{\lim }}\,x\ln x=0.$

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