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clara1223

  • one year ago

find the limit as x approaches 121 of (sqrt(x)-11)/(x-121). I have already solved and gotten 1/22 which was right but I can't remember how and can't find my notes. All I need is an explanation

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  1. clara1223
    • one year ago
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    \[\lim_{x \rightarrow 121}\frac{ \sqrt{x}-11 }{ x-121 }\]

  2. amistre64
    • one year ago
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    can we factor the bottom?

  3. clara1223
    • one year ago
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    if x was squared then yes but it's not.

  4. amistre64
    • one year ago
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    oh but it is ... let a^2 = x, then a = sqrt(x)

  5. clara1223
    • one year ago
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    wait I don't understand that, can you explain it further?

  6. amistre64
    • one year ago
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    also, as a general rule, if the top and bottom have the same zero ... the zeros cancel out due to a common factor.

  7. amistre64
    • one year ago
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    at x=11, we have 0/0 right?

  8. clara1223
    • one year ago
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    do you mean at x=121?

  9. amistre64
    • one year ago
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    its just a substitution ... if the x does not look like its a square ... replace it by something that looks squarey yeah, x = 121 ... brain is faster then my fingers sometimes

  10. clara1223
    • one year ago
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    how would you go about replacing it with something that looks "squarey"?

  11. amistre64
    • one year ago
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    if we let x=a^2, and take the sqrt of each side ... sqrt(x) = a a-11 --------- a^2 - 121 now it looks squared to me

  12. clara1223
    • one year ago
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    Oh! that helps a lot. So then can factor the denominator to (a+11)(a-11) and cancel a-11 from both sides so im left with 1/a+11?

  13. amistre64
    • one year ago
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    your notes might say; divide it al by the highest power of x tho

  14. amistre64
    • one year ago
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    yes

  15. amistre64
    • one year ago
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    and since sqrt(x) = a sqrt(121) = 11 and we get your coveted 1/22

  16. clara1223
    • one year ago
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    Thank you!

  17. amistre64
    • one year ago
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    factoring seems to be the best route to me ... and youre welcome again :)

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