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muffin

  • 2 years ago

Evaluate the limit, if it exists lim h-0 ((1/(x+h)^2 - 1/x^2 ))/ h

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  1. myininaya
    • 2 years ago
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    Try combining top fractions. And get rid of the compound fraction. Unless you already know derivatives then we can actually skip this part.

  2. muffin
    • 2 years ago
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    so a common denominator for the top fractions? do i expand the (x+h)^2

  3. myininaya
    • 2 years ago
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    You will need to.

  4. muffin
    • 2 years ago
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    so x^2 + 2xh + h^2, what would be the common denominator

  5. myininaya
    • 2 years ago
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    x^2(x+h)^2

  6. myininaya
    • 2 years ago
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    |dw:1379983226588:dw|

  7. myininaya
    • 2 years ago
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    |dw:1379983241382:dw|

  8. muffin
    • 2 years ago
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    2xh + h^2/ x^2 (x+h)^2 x (1/h)

  9. muffin
    • 2 years ago
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    can i factor out a h from the numerator?

  10. muffin
    • 2 years ago
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    so i factored out an h from the numerator, than cancelled the h from 1/h

  11. myininaya
    • 2 years ago
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    What happen to your negative in from 2xh?

  12. myininaya
    • 2 years ago
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    h/h=1 yep yep

  13. muffin
    • 2 years ago
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    i have lim h--0 2x+h/ x^2(x+h)^2

  14. muffin
    • 2 years ago
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    so now i can "sub" in 0 into h, but what value goes for x?

  15. myininaya
    • 2 years ago
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    Well yeah but still you are missing a negative on top

  16. myininaya
    • 2 years ago
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    only h is going somewhere x stays

  17. myininaya
    • 2 years ago
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    It said what happens as h goes to zero (it said nothing for x)

  18. myininaya
    • 2 years ago
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    Did you figure out what negative you are missing?

  19. muffin
    • 2 years ago
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    ok i think i see my mistake so i went thru the stages again and right now I have lim h--0 -2x-h/x^2 (x+h)^2

  20. muffin
    • 2 years ago
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    because I factored out an h, but im confused as to my next step

  21. myininaya
    • 2 years ago
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    h goes to 0

  22. muffin
    • 2 years ago
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    is my answer suppose to be just a number? or will it have a value and x

  23. myininaya
    • 2 years ago
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    Your answer will be a function of x.

  24. muffin
    • 2 years ago
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    -2x-0/x^2 (x+0) ^2 ?

  25. myininaya
    • 2 years ago
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    (-2x-0)/(x^2(x+0)^2) yes Simplify.

  26. muffin
    • 2 years ago
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    -2/x?

  27. muffin
    • 2 years ago
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    not rlly syre how to simplify the denominator

  28. mathslover
    • 2 years ago
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    \(\cfrac{(-2x-0)}{(x^2)(x+0)^2}\) (Just for ease - wrote the LaTeX)

  29. myininaya
    • 2 years ago
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    -2x-0=-2x x^2(x+0)^2=x^2(x)^2 (since x+0=x) Use law of exponents.

  30. myininaya
    • 2 years ago
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    Recall one of the laws of exponents a^m * a^n=a^(m+n)

  31. muffin
    • 2 years ago
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    ok so my answer is -2/x^3

  32. myininaya
    • 2 years ago
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    yep

  33. muffin
    • 2 years ago
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    thank you :)

  34. mathslover
    • 2 years ago
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    But the denominator should be x^4 @myin

  35. myininaya
    • 2 years ago
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    -2x/x^4=-2/x^3

  36. muffin
    • 2 years ago
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    I have another question: find the limit if it exists limx-- -2 2-|x|/ 2+x

  37. muffin
    • 2 years ago
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    thats lim x approaches -2

  38. muffin
    • 2 years ago
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    can I just sub in -2 for x? so that makes it 0/0

  39. mathslover
    • 2 years ago
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    Oh sorry @myininaya - didn't saw that :) Thanks !

  40. myininaya
    • 2 years ago
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    \[\lim_{x \rightarrow -2} \frac{2-|x|}{2+x}\] ?

  41. muffin
    • 2 years ago
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    yes

  42. myininaya
    • 2 years ago
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    So we can't just simply plug in -2 because our denominator will be 0. Is there someway to make the function continuous at x=-2 so we can just plug in Well |x|=x if x>=0 and -x if x<0 Since x approaches -2 then x is negative since we are looking at what surrounds a negative number so we have |x|=-x

  43. myininaya
    • 2 years ago
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    That is your key ingredient

  44. muffin
    • 2 years ago
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    okay

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