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Is the square root of infinity still infinity?
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yes
Yes it is
ya
or rather, it is undefined, or indeterminate
there are different sizes of infinity ...
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if you take the square root of a very very very big number , you will still have a very very very big number , now if you make the number even bigger , the square root will also get bigger , now imagine what will happen if this number is infinite..
\[\lim_{n\to \infty}\sqrt{n} = \infty \] yes
|dw:1330637669831:dw| so I'm doing a definite integral, and 2 times square root of infinity is part of my answer; does that mean it diverges?
yes
a definite integral is the limit as you approach a bad value ...
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if the infinity is in a denominator it goes to 0; if its not under the bar, it goes to infinity
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