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Dodo1 Group Title

-infity and infity, whats the difference?

  • one year ago
  • one year ago

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  1. Dodo1 Group Title
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    \[\infty -\infty\]\[\frac{ Sqrt(x^2+14x) }{ 14-13x } \]

    • one year ago
  2. ghazi Group Title
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    answer of your first question, what is the difference between - infinity and +infinity, lets say if i ask you to divide a line segment in smaller parts , you will keep chopping it off till it is visible to you but if we asked how smaller we can chop, the answer is infinity and if i ask you to imagine how far you can go in the universe the answer is again infinity, in terms of mathematics when you go in the negative direction of a number line then you tend to approach - infinity and when you go in + direction you approach +infinity

    • one year ago
  3. ghazi Group Title
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    and basically, all the numbers lie between -infinity and +infinity

    • one year ago
  4. ghazi Group Title
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    and i dont know what is the limit of the question that you have given :(

    • one year ago
  5. Dodo1 Group Title
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    how do you evaluate the limit \[\lim_{x \rightarrow \infty }\sqrt{x^2+3x+1}-x\]

    • one year ago
  6. Dodo1 Group Title
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    and thank you for the explantion. but is it really matter ( - or + infitiy) when evaluate the limit?

    • one year ago
  7. ghazi Group Title
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    of course it does matter where your function is moving, and if you just see the function that you have given, you can answer it with logic and theoretical explanation that is, when x is tending to infinity the value of your function would be infinite just substitute a huge number and see what you get and if you are getting infinity-infinity then that is kinda undetermined form

    • one year ago
  8. Dodo1 Group Title
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    I see, how do i do the first question>?

    • one year ago
  9. ghazi Group Title
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    \[\sqrt{(x^2+3x+1)}-x * \frac{ x+\sqrt{(x^2+3x+1)} }{ x+ \sqrt{x^2+3x+1} }\] simplify this and your answer would be 3/2

    • one year ago
  10. Dodo1 Group Title
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    what about negeative infity?

    • one year ago
  11. Dodo1 Group Title
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    the answer is not the same isnt it?

    • one year ago
  12. ghazi Group Title
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    well 3/2 is the answer of \[\lim x> \infty \sqrt{x^2+3x+1}-x\]

    • one year ago
  13. ghazi Group Title
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    and the answer of your first question is infinity because just use L' hospital's rule and see after differentiating you will get 2x/-13 now if you're going to substitute infinity at place of x then you will have infinity as answer too

    • one year ago
  14. Dodo1 Group Title
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    thank you very much!

    • one year ago
  15. ghazi Group Title
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    :D

    • one year ago
  16. Dodo1 Group Title
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    medels *infinity for you!

    • one year ago
  17. ghazi Group Title
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    thank you very much :D :)

    • one year ago
  18. SithsAndGiggles Group Title
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    Another way of finding the limit: \[\lim_{x\to-\infty}\left(\sqrt{x^2+3x+1}-x\right)\\ \lim_{x\to-\infty}\left(\sqrt{x^2}\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}-x\right)\\ \lim_{x\to-\infty}\left(|x|\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}-x\right)\\ \text{Since $x\to-\infty, |x|=-x$. If that doesn't make sense,}\\ \text{look up the definition of absolute value.}\\ \lim_{x\to-\infty}\left(-x\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}-x\right)\\ \lim_{x\to-\infty}(-x)\left(\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}+1\right)=-(-\infty)(\sqrt1+1)=\infty\]

    • one year ago
  19. SithsAndGiggles Group Title
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    \[\lim_{x\to\infty}\frac{\sqrt{x^2+14x}}{14-13x}\\ \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{1+\frac{14}{x}}}{14-13x}\\ \lim_{x\to\infty}\frac{|x|\sqrt{1+\frac{14}{x}}}{14-13x}\\ \text{Since $x\to\infty,|x|=x$.}\\ \lim_{x\to\infty}\frac{x\sqrt{1+\frac{14}{x}}}{14-13x}\\ \lim_{x\to\infty}\frac{\sqrt{1+\frac{14}{x}}}{\frac{14}{x}-13}=\frac{\sqrt1}{-13}=-\frac{1}{13}\]

    • one year ago
  20. Dodo1 Group Title
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    thats really cool too.

    • one year ago
  21. Dodo1 Group Title
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    but why its infy(sqrt1)+1=infy ?

    • one year ago
  22. SithsAndGiggles Group Title
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    \[-(-\infty)\left(\sqrt1+1\right)=\infty\left(1+1\right)=\infty(2)=\infty\]

    • one year ago
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