clara1223
  • clara1223
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
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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clara1223
  • clara1223
\[\lim_{x \rightarrow 121}\frac{ \sqrt{x}-11 }{ x-121 }\]
amistre64
  • amistre64
can we factor the bottom?
clara1223
  • clara1223
if x was squared then yes but it's not.

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amistre64
  • amistre64
oh but it is ... let a^2 = x, then a = sqrt(x)
clara1223
  • clara1223
wait I don't understand that, can you explain it further?
amistre64
  • amistre64
also, as a general rule, if the top and bottom have the same zero ... the zeros cancel out due to a common factor.
amistre64
  • amistre64
at x=11, we have 0/0 right?
clara1223
  • clara1223
do you mean at x=121?
amistre64
  • amistre64
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
clara1223
  • clara1223
how would you go about replacing it with something that looks "squarey"?
amistre64
  • amistre64
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
clara1223
  • clara1223
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?
amistre64
  • amistre64
your notes might say; divide it al by the highest power of x tho
amistre64
  • amistre64
yes
amistre64
  • amistre64
and since sqrt(x) = a sqrt(121) = 11 and we get your coveted 1/22
clara1223
  • clara1223
Thank you!
amistre64
  • amistre64
factoring seems to be the best route to me ... and youre welcome again :)

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