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Christos

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

  • 11 months ago
  • 11 months ago

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  1. Christos
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    what...?

    • 11 months ago
  2. Christos
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    @ajprincess

    • 11 months ago
  3. Christos
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    @LoveYou*69

    • 11 months ago
  4. Christos
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    @ganeshie8

    • 11 months ago
  5. ajprincess
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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

    • 11 months ago
  6. ajprincess
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    Sorry if I have confused u

    • 11 months ago
  7. BrLima
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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.

    • 11 months ago
  8. Christos
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    6 or lim6 ?

    • 11 months ago
  9. ajprincess
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    jst 6.

    • 11 months ago
  10. ajprincess
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    |dw:1367516895165:dw|

    • 11 months ago
  11. Christos
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    thx man

    • 11 months ago
  12. ajprincess
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    welcome:) do u get it?

    • 11 months ago
  13. Christos
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    How do I solve lim(x--->1) (x^4-1)/(x-1) it gives me 0

    • 11 months ago
  14. ajprincess
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    nw try to simplify it as possible.

    • 11 months ago
  15. Christos
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    Please give me steps, just on this one, then I will get it

    • 11 months ago
  16. ajprincess
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    \(x^4-1=(x^2)^2-(1^2)^2=(x^2-1^2)(x^2+1^2)=(x-1)(x+1)(x^2+1)\) \[\frac{x^4-1}{x-1}=\frac{(x-1)(x+1)(x^2+1)}{x-1}\] \[=?\]

    • 11 months ago
  17. Christos
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    1

    • 11 months ago
  18. ajprincess
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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

    • 11 months ago
  19. Christos
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    How did you get from here (x2−12)(x2+12) to here (x−1)(x+1)(x2+1)

    • 11 months ago
  20. ajprincess
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    \((x^2-1^2)=(x-1)(x+1)\) difference of squares. so \((x^2-1^2)(x^2+1)=(x-1)(x+1)(x^2+1)\)

    • 11 months ago
  21. Christos
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    its 3??

    • 11 months ago
  22. ajprincess
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    \((x+1)(x^2+1)\) when x=1 \((1+1)(1+1)\) \(=2*2\)

    • 11 months ago
  23. Christos
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    I see!

    • 11 months ago
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