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Christos Group Title

Help me solve lim(x-->2) x(x-1)(x+1) I merely wish to understand the logic behind it

  • one year ago
  • one year ago

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  1. Christos Group Title
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    what...?

    • one year ago
  2. Christos Group Title
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    @ajprincess

    • one year ago
  3. Christos Group Title
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    @LoveYou*69

    • one year ago
  4. Christos Group Title
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    @ganeshie8

    • one year ago
  5. ajprincess Group Title
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    first plug in the value given as the value that x tends to in place of x in given limit. in this case it is given that x->2. so plug in 2 in x(x-1)(x+1) and find the value. in some cases when u plug in the values that x tends to, u get zero. in that cases u try to simplify as much as possible and then plug in the value and simplify. Let me give u an example for such cases. lim x^2-9/x-3 x->3 when u plug in x=3 u get (3)^2-9/3-3=9-9/0=0/0-indeterminate. x^2-9/x-3=(x-3)(x+3)/(x-3)=x+3 nw plug in 3 , u get 3+3=6. so that is the answer for question lim x^2-9/x-3 x->3. Does that help? @christos

    • one year ago
  6. ajprincess Group Title
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    Sorry if I have confused u

    • one year ago
  7. BrLima Group Title
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    Yes, the answers is 6. The way to find it was described by @ajprincess. When the function is continuous you just have to plug in the value. However, if the function isn't continous on the specific point you'll try to simplify the function. After the simplification you'll have a new function that is continuous, so you can just plug in the value to find the limit.

    • one year ago
  8. Christos Group Title
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    6 or lim6 ?

    • one year ago
  9. ajprincess Group Title
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    jst 6.

    • one year ago
  10. ajprincess Group Title
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    |dw:1367516895165:dw|

    • one year ago
  11. Christos Group Title
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    thx man

    • one year ago
  12. ajprincess Group Title
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    welcome:) do u get it?

    • one year ago
  13. Christos Group Title
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    How do I solve lim(x--->1) (x^4-1)/(x-1) it gives me 0

    • one year ago
  14. ajprincess Group Title
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    nw try to simplify it as possible.

    • one year ago
  15. Christos Group Title
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    Please give me steps, just on this one, then I will get it

    • one year ago
  16. ajprincess Group Title
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    \(x^4-1=(x^2)^2-(1^2)^2=(x^2-1^2)(x^2+1^2)=(x-1)(x+1)(x^2+1)\) \[\frac{x^4-1}{x-1}=\frac{(x-1)(x+1)(x^2+1)}{x-1}\] \[=?\]

    • one year ago
  17. Christos Group Title
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    1

    • one year ago
  18. ajprincess Group Title
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    \[\frac{(x-1)(x+1)(x^2+1)}{x-1}=\frac{\cancel{(x-1)}(x+1)(x^2+1)}{\cancel{x-1}}\] \[=(x+1)(x^2+1)\] nw plug in 1 in place of x nd find the value

    • one year ago
  19. Christos Group Title
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    How did you get from here (x2−12)(x2+12) to here (x−1)(x+1)(x2+1)

    • one year ago
  20. ajprincess Group Title
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    \((x^2-1^2)=(x-1)(x+1)\) difference of squares. so \((x^2-1^2)(x^2+1)=(x-1)(x+1)(x^2+1)\)

    • one year ago
  21. Christos Group Title
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    its 3??

    • one year ago
  22. ajprincess Group Title
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    \((x+1)(x^2+1)\) when x=1 \((1+1)(1+1)\) \(=2*2\)

    • one year ago
  23. Christos Group Title
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    I see!

    • one year ago
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