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Babynini

  • one year ago

Limits....again. Help!

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  1. Babynini
    • one year ago
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    Both numbers 6 and 7, but let's begin with 6!!

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  2. Babynini
    • one year ago
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    @peachpi @zepdrix :)

  3. Astrophysics
    • one year ago
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    L'hospital for first one

  4. Babynini
    • one year ago
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    #6) b = 4?

  5. zepdrix
    • one year ago
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    Ya :) Imma wait and see what Ganeshie has to say though ^^

  6. ganeshie8
    • one year ago
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    Notice that the denominator is 0 when you plugin x=0 Also, for the limit to exist (finite), you must get the indeterminate 0/0 form when you plugin x=0. otherwise the limit does not exist (infinite). so, the numerator part must evaluate to 0 at x=0

  7. ganeshie8
    • one year ago
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    \(\sqrt{ax+b}-2\) evaluating it at \(x=0\) and setting equal to \(0\) does give you \(b=4\)

  8. ganeshie8
    • one year ago
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    plugin the \(b\) value, rationalize and you will see what \(a\) needs to be

  9. Astrophysics
    • one year ago
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    For 7 note this |dw:1444533414414:dw| you can do the same with |2x-1| you'll have an ugly numerator at the beginning, but it'll simplify to something nice :D

  10. Babynini
    • one year ago
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    Right, so I got b. But the a is being something terrible.

  11. ganeshie8
    • one year ago
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    \[\frac{\sqrt{ax+4}-2}{x}=\frac{ax+4-4}{x(\sqrt{ax+4}+2)} = \frac{a}{\sqrt{ax+4}+2}\] plugin x=0 and you will see immediately that \(a\) needs to be \(4\)

  12. dan815
    • one year ago
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    |dw:1444533867422:dw|

  13. Babynini
    • one year ago
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    @ganeshie8 thanks! I got it :)

  14. Babynini
    • one year ago
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    @dan815 gwow you are doing some fancy stuff, I think I also follow wht you are doing :D

  15. Babynini
    • one year ago
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    mm well if a = 4 and b = 4..then it works.

  16. Babynini
    • one year ago
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    but it's always over 0 :/

  17. Astrophysics
    • one year ago
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    a should be 4

  18. Babynini
    • one year ago
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    Ok, yeah that's what I got xD thanks yall :)

  19. dan815
    • one year ago
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    you actully do get a limit!

  20. dan815
    • one year ago
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    |dw:1444534732654:dw|

  21. dan815
    • one year ago
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    okay i hope thats clear now

  22. dan815
    • one year ago
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    i think this is how lhopital is justified too

  23. dan815
    • one year ago
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    we are looking at the behavior of your function very locally, so we'd only care about the slopes changes around that point

  24. Babynini
    • one year ago
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    Righto, 6 all makes sense :) I'm going to start a new thread for number 7

  25. dan815
    • one year ago
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    so for example the lhopital way d/x(abs(2x-1)- abs(2x+)) --------------------- d/dx(x) d/dx ( abs(2x-1) - d/dx(abs(2x+1)) ------------------------------- 1 now derivative of abs(2x-1) near 0 = is -2 and the detiaive of abs(2x+1) near 0 is 2 so -2 - 2 ------- 1 =-4 as x-->0

  26. Babynini
    • one year ago
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    Em, haven't learnt derivatives yet :/

  27. dan815
    • one year ago
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    okay then do it the other way

  28. Astrophysics
    • one year ago
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    I think multiplying by conjugate might work

  29. dan815
    • one year ago
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    yeah

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