What is the required limit of lim -0+ ...(sin(x))^x ?

Question

What is the required limit of lim ->0+ ...(sin(x))^x ?

anonymous · Answer

uh, what?

$$\lim_{x \to 0}+\dots \sin^x(x)$$?

konradzuse · Answer

Yeah it's asking "what is the required limit"  I have no idea what they are asking...

anonymous · Answer

I have no idea what the expression even means. It isn't written using (standard) legible notation.

konradzuse · Answer

Without the .... obviously...  I was doing that so it didn't look like 0+ (sin(x))^x

konradzuse · Answer

$$\lim_{x \rightarrow 0+} (\sin(x))^x$$

anonymous · Answer

That's more clear. That's actually solvable. I think it's asking for the right-hand limit.

turingtest · Answer

I agree^
LaTeX note:  \lim_{x\to0^+}
$$ \lim_{x\to0^+}$$that makes it more clear :)

turingtest · Answer

is there a better way to do this than doing the exponent trick?

anonymous · Answer

I'm confused as to why we can't evaluate this simply.
$$\lim_{x \to 0^{+}}\sin^x(x)=sin^0(0)=(0)^0=1?$$

turingtest · Answer

0^0 is undefined

konradzuse · Answer

I got that, but I'm not sure what they mean by the "limit." and what we are trying to solve for...

anonymous · Answer

@TuringTest, no. It is conventionally defined. Most authors take $0^0=1$.

turingtest · Answer

Except for by some mathematicians like Euler who took it upon themselves to define 0^0=1 I beg to differ. The truth is that the issue is still very debated.
But don't take my word for it, let me find a good link I saw once, one sec...

turingtest · Answer

http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/#b

anonymous · Answer

I have read about it. It's generally a convention/definition. See http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/

anonymous · Answer

But we are detracting from the question here. Is that the actual answer?

turingtest · Answer

I dunno, let's just exponentiate

konradzuse · Answer

yes I just ented 1 and i was correct.

konradzuse · Answer

I got that before, but I dind't understand exactly what they were asking... Solve for the limit wuld havebeen simpler..  I thought they were asking to change the imit to something else which made no sense.. Thanks Limitless...

anonymous · Answer

You're welcome, @KonradZuse! :) Also, this is freaky: http://www.wolframalpha.com/input/?i=lim+as+x+to+0%5E%2B+of+sin%5Ex%28x%29

konradzuse · Answer

http://www.wolframalpha.com/input/?i=limit+x+-%3E+0%2B++of+%28sin%28x%29%29%5Ex This is what I had :P

konradzuse · Answer

Pool time everyone, ttyl.

anonymous · Answer

@KonradZuse, ah! WolframAlpha failed to interpret it. Nice catch.