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appleduardo

  • 3 years ago

what is the limit for the next function?

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  1. appleduardo
    • 3 years ago
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    \[\frac{ x ^{4} -16 }{x ^{2} -2 }\]

  2. appleduardo
    • 3 years ago
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    x--> 2

  3. hartnn
    • 3 years ago
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    can u factor x^4 -16 ?

  4. hartnn
    • 3 years ago
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    wait, is that x^2 -2 ?? if yes, there is no need to factor. just put x=2

  5. appleduardo
    • 3 years ago
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    mm maybe do this: x[x^(3) ] +16 ooh i was wrong, is x^(2) - 4

  6. hartnn
    • 3 years ago
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    x^4 -16 now 16 is 4^2 , isn't it ? so you can write numerator as \(\large (x^2)^2-4^2\) use a^2-b^2 = .... formula, do u know ?

  7. appleduardo
    • 3 years ago
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    mm i dont remember! but once i have the numerator like that what can i do next?

  8. hartnn
    • 3 years ago
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    \(\large a^2-b^2=(a+b)(a-b) \\ \text {so what is} (x^2)^2-4^2 \)

  9. appleduardo
    • 3 years ago
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    so i'll get (x+2)^(2) * ( x-2)^(2) ???

  10. hartnn
    • 3 years ago
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    ?? no. \(\huge (x^2)^2-4^2=(x^2+4)(x^2-4)\)

  11. appleduardo
    • 3 years ago
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    ooh!! i understand now ! :D so then i can cancel x^(2) - 4 and the limit will be 8 right?

  12. hartnn
    • 3 years ago
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    YES.

  13. appleduardo
    • 3 years ago
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    Wow! you're awesome !! :D mm what about this one? \[\frac{ \frac{ 1 }{ x } + \frac{ 1 }{ 2 }}{ x-2 }\] x-->2

  14. hartnn
    • 3 years ago
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    that limit does not exist!

  15. appleduardo
    • 3 years ago
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    mm why?

  16. hartnn
    • 3 years ago
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    u know the how to find lim x->2+ and lim x->2- or know what they are ?

  17. appleduardo
    • 3 years ago
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    mm i dont :/

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