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Christos

  • one year ago

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

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  1. Christos
    • one year ago
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    what...?

  2. Christos
    • one year ago
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    @ajprincess

  3. Christos
    • one year ago
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    @LoveYou*69

  4. Christos
    • one year ago
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    @ganeshie8

  5. ajprincess
    • one year ago
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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

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

  7. BrLima
    • one year ago
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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.

  8. Christos
    • one year ago
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    6 or lim6 ?

  9. ajprincess
    • one year ago
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    jst 6.

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

  11. Christos
    • one year ago
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    thx man

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

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

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

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

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

  17. Christos
    • one year ago
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    1

  18. ajprincess
    • one year ago
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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

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

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

  21. Christos
    • one year ago
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    its 3??

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

  23. Christos
    • one year ago
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    I see!

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