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anonymous

  • one year ago

How do you find the limit when x is approaching 0, (x^2+3)/x^4? @welshfella

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  1. anonymous
    • one year ago
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    @Phi

  2. anonymous
    • one year ago
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    @zepdrix anyone out there

  3. anonymous
    • one year ago
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    You could graph it and look, that's always a good place to start

  4. anonymous
    • one year ago
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    @RunawayGalaxy I honestly don't even know how to graph that.

  5. zepdrix
    • one year ago
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    so ummmmmm :)

  6. zepdrix
    • one year ago
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    \[\large\rm \lim_{x\to0}\frac{x^2+3}{x^4}\]Simply plugging in 0 gives you an indeterminate form of \(\large\rm \frac{3}{0}\) ya?

  7. anonymous
    • one year ago
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    @zepdrix yes

  8. zepdrix
    • one year ago
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    Do you remember anything about this limit?\[\large\rm \lim_{x\to0}\frac{1}{x}\]Because our problem is behaving the same way.

  9. anonymous
    • one year ago
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    I do not remember much about that particular limit. Would I replace x with 0? 1/0 is undetermined?

  10. zepdrix
    • one year ago
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    Well let's plug in a value that's very close to 0, that should give us an idea of what's going on. How about 1/99999, that's pretty close to 0 right? :)\[\large\rm \lim_{x\to0}\frac{1}{\color{orangered}{x}}\approx\frac{1}{\color{orangered}{\frac{1}{99999}}}=1\cdot\frac{99999}{1}=99999\]Do you understand what I did here? I plugged in a really really small fraction, a value close to zero. And applying an algebra rule, the result is giving us a really big value.

  11. zepdrix
    • one year ago
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    So as x is getting closer to 0, 1/x is growing infinitely large.

  12. anonymous
    • one year ago
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    @zepdrix Does L'hopital's rule not apply in this case? Is 3/0 not indeterminate form?

  13. zepdrix
    • one year ago
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    It's not the indeterminate form that we need for L'Hop unfortunately :( Need 0/0 or infty/infty

  14. anonymous
    • one year ago
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    I remembered that there were a bunch, but couldn't figure if that was one of them

  15. zepdrix
    • one year ago
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    @raesal what do you think? :d do you understand why 1/x is blowing up large and in charge like that? :D make sense of the example?

  16. anonymous
    • one year ago
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    @zepdrix I understand the example, I just don't understand how that example is going to help me solve this problem. I don't understand how 1/x "blowing up large and in charge" will help me solve this problem.

  17. zepdrix
    • one year ago
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    Well it's blowing up like, is partly explained by the fact that it's approaching the indeterminate form 1/0. Notice that with our problem, we're approaching the indeterminate form 3/0. So we will end up with the same result. I'm not sure of the exact algebra steps.. umm It's easier to think about it intuitively. The bottom is a 4th power, so it's doing everything "faster" than the top. It's getting to zero faster than the numerator, so it blows up just like 1/x. Oh oh maybe we could do this: Multiply top and bottom by 1/x^4\[\large\rm \lim_{x\to0}\frac{x^2+3}{x^4}\left(\frac{1/x^4}{1/x^4}\right)=\lim_{x\to0}\frac{\frac{1}{x^2}+\frac{3}{x^4}}{1}=\lim_{x\to0}\frac{1}{x^2}+\lim_{x\to0}\frac{3}{x^4}\]

  18. zepdrix
    • one year ago
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    Then we can relate it back to that 1/x limit in each of these limits,\[\large\rm =\left(\lim_{x\to0}\frac{1}{x}\right)^2+3\left(\lim_{x\to0}\frac{1}{x}\right)^4\]

  19. zepdrix
    • one year ago
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    That seems wayyy tedious though :d Intuitionnnnnnn instead!

  20. zepdrix
    • one year ago
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    That multiplication step was kinda sloppy, i shoulda just separated the fractions, x^2/x^4 and 3/x^4, same result.

  21. zepdrix
    • one year ago
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    I dunno, it's a lot to take in >.< Lemme know what you're thinkin

  22. anonymous
    • one year ago
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    I'm confused. I'm so sorry >.<

  23. zepdrix
    • one year ago
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    When you have a fraction:\[\large\rm \lim_{x\to 0}\frac{stuff}{other~stuff}\] If the stuff in the numerator is going towards 0 faster, then the fraction overall is going towards zero. It's approaching this form: \(\large\rm \frac{0}{number}\) which is just zero. When the bottom is going towards 0 faster, then the whole thing is approaching infinity. It's approaching this form: \(\large\rm \frac{number}{0}\) This second case is what is happening with our problem, the bottom is getting to zero much much faster.

  24. zepdrix
    • one year ago
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    Blah, math is hard >.<

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