What does Almost always mean, and why is this result true? I've seen examples of when this works, of course, but why should it "almost always" work?
Show that $\lim_{x,y\to 0^+} x^y$ is "almost always" equal to $1$
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real-analysis
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2I think you want to take the limit as $x\rightarrow0^+$? – 2012-12-10
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0@DavidMitra Yes, that's exactly what I meant, it was a mistype. Wow, already got 2 downvotes in under 2 minutes. Tad bit harsh. – 2012-12-10
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0I agree. I think if you provide a bit more information concerning who said it's "almost always equal to 1" and in what context it was said, people wouldn't jump the gun with downvotes. As the question is phrased, saying 'almost always' is nonsense. The limit is $1$; end of story... (Note please you want the limit *from the right*, that is as $x\rightarrow 0^{\color{red}+}$ though.) – 2012-12-10
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0Right, I was mistaken twice. I've edited the question again. This is exactly what it said, as it is written now. – 2012-12-10
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0Aha! You snuck a $y$ in the limit. Now the question, and the phrase "almost always", makes more sense. – 2012-12-10
1 Answers
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If $(x,y)$ approaches $(0,0)$ from within a sector in the first quadrant bounded by lines $y=ax$ and $y=bx$, then the limit of $x^y$ is $1$. You can show that by having it first approach along those two lines, and then squeezing. So it can approach a limit other than $1$ only if it follows a path having either the $x$-axis or the $y$-axis as a tangent line.