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DLS

  • 3 years ago

Limits

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  1. DLS
    • 3 years ago
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    \[\lim_{x \rightarrow \infty }\frac{x^4+2x^3+3}{2x^4-x+2}\]

  2. DLS
    • 3 years ago
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    x+1 is a factor of this right?

  3. DLS
    • 3 years ago
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    however,no use..

  4. TomLikesPhysics
    • 3 years ago
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    I think it is one half. Because you got up- and downstairs and x^4 which I think is a special case and in which the limit approaches the coefficents before these two identical powers of x which is 1upstairs and 2 downstairs - 1/2.

  5. DLS
    • 3 years ago
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    it is 1/2 only but how

  6. TomLikesPhysics
    • 3 years ago
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    If you would use the L`Hopital Rule you would end up with it.

  7. TomLikesPhysics
    • 3 years ago
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    Do you want to see the calculation?

  8. DLS
    • 3 years ago
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    \[\lim_{x \rightarrow \infty }\frac{3x^4+4x^3+}{4x^4+x^2+7}\] I'm trying to do this with the same concept but i'm getting 5/7 @TomLikesPhysics ?

  9. DLS
    • 3 years ago
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    +5

  10. DLS
    • 3 years ago
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    *

  11. DLS
    • 3 years ago
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    \[\frac{12x^3+12x^2}{16x^3+2x}\]

  12. TomLikesPhysics
    • 3 years ago
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    should be 3/4

  13. DLS
    • 3 years ago
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    ??

  14. TomLikesPhysics
    • 3 years ago
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    the one where you got 5/7

  15. DLS
    • 3 years ago
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    okay got it!

  16. DLS
    • 3 years ago
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    but how is it gaining infinity/infinity?

  17. TomLikesPhysics
    • 3 years ago
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    How it gains infinity? You take the limit where x goes to infinity. So both terms grow. Is that what you meant?

  18. DLS
    • 3 years ago
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    it is something+infinity is it samething?

  19. DLS
    • 3 years ago
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    \[\lim_{x \rightarrow \infty } \frac{x^3+x^2+4}{x^2+8}\] are you gettng 1 here?

  20. TomLikesPhysics
    • 3 years ago
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    Here it should be infinity. You got downstairs an x^2 and upstairs and x^3. So the x^3 "is stronger" and wins and therefore the whole thing goes to infinity.

  21. DLS
    • 3 years ago
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    No,I solved according to what you said and got this: \[\frac{3x^2+2x}{2x}\] so I got 3x+1=1?

  22. TomLikesPhysics
    • 3 years ago
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    How do you end up with 3x+1=1 if you got a fraction?

  23. DLS
    • 3 years ago
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    \[\frac{3x^2}{2x} + \frac{2x}{2x}\]

  24. DLS
    • 3 years ago
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    I ended up with: \[\frac{3}{2}x+1\]

  25. TomLikesPhysics
    • 3 years ago
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    k but if you got 3x+1 and let x approach infinity (taking the limit) than you end up with infinity.

  26. DLS
    • 3 years ago
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    okay! if it was 1/x and x==> infinity it would have been 1 RIGHT?

  27. TomLikesPhysics
    • 3 years ago
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    limit x-> inf. of ax+b is alsways infinity. The b does not matter in the long run and nether does a while x gets bigger and bigger.

  28. DLS
    • 3 years ago
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    Great.Clear now :)

  29. TomLikesPhysics
    • 3 years ago
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    If you got 1/x and x gets bigger and bigger you would get a limit which is 0.

  30. DLS
    • 3 years ago
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    yeah!

  31. TomLikesPhysics
    • 3 years ago
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    1/x as x approaches infinity is 0. 1/x as x approaches 0 is infinity.

  32. DLS
    • 3 years ago
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    \[\lim_{x \rightarrow 0} \frac{\cos7x-\cos9x}{\cos3x-\cos5x}\] These ones?

  33. TomLikesPhysics
    • 3 years ago
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    I don´t know. Have not seen this one before.^^ Never dealt with limits and sine or cosine.

  34. DLS
    • 3 years ago
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    okay..book did a direct step after this :| \[\frac{2\sin8xsinx}{2\sin4xsinx} \]

  35. TomLikesPhysics
    • 3 years ago
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    ^^ First time I see that kind of stuff. I´m sorry but with that I can not help you.

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