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This is not a homework question, but it's similar, I'm studying for my exam. And I'm sorry for asking, but I can't put a dent in it for over a day. I keep coming back to it, and I don't know what to do.
So, I need to show this function is continuous at (1,0), and WolframAlpha shows it is (http://www.wolframalpha.com/input/?i=Limit%5B%5B%2F%2Fmath:ln%5E3(x)%2F(sqrt((x-1)%5E2%2By%5E4)-1)%2F%2F%5D,+%7B%5B%2F%2Fmath:x,y%2F%2F%5D%7D+-%3E+%7B%5B%2F%2Fmath:1,+0%2F%2F%5D%7D%5D).
Before even trying WA, I tried to disprove that the limit exists, at x=1, y=0 and x-1=y, and I keep getting zero. The only way I know of proving a limit exists is a squeeze th. And that's where I can't move further from 0<=abs(f(x,y)), and then recognizing that the denominator is >=0. That's it, I have no idea what to do next. I beg of you, any suggestions would be much appreciated.
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Also, for what is worth, I've been googling for ages. And any squeeze th. examples online are just way to easy. I have no CHANCE of getting something like that on my exam, our professor is a complete sadist.
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Please guys, people who are good at calc III are very scarce, I have literally no one to ask, all I find online is obvious stuff.
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Could I expand (x-1)^2 to get x^2-2x+1 and then get rid of x^2 and y^4?
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try killing urself
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>>8970906
That's not very nice.
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Okay I've rationalized the god damn denominator, and I don't think it did anything. I'm swamped, fuck it at this point honestly
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directly plugging in x=1 and y=0 yields 0/-1 = 0, no?
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>>8970944
It actually does and thank you so much for trying, but I fucked up the link... This is the actual limit is (http://www.wolframalpha.com/input/?i=Limit%5B%5B%2F%2Fmath:ln%5E3(x)%2F(sqrt((x-1)%5E2%2By%5E4%2B1)-1)%2F%2F%5D,+%7B%5B%2F%2Fmath:x,y%2F%2F%5D%7D+-%3E+%7B%5B%2F%2Fmath:1,+0%2F%2F%5D%7D%5D).
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