jpsmarinho
  • jpsmarinho
lim sqrt(x^2-9)/(2x-6) when x goes to -infinity. I use a software and discover that the lim equals to -1/2, but, when I do, II only encounter the result 1/2
Mathematics
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jpsmarinho
  • jpsmarinho
lim sqrt(x^2-9)/(2x-6) when x goes to -infinity. I use a software and discover that the lim equals to -1/2, but, when I do, II only encounter the result 1/2
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
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jpsmarinho
  • jpsmarinho
Ah but has square root in the x^2 -9
jpsmarinho
  • jpsmarinho
The lim is the following: {(x^2-9)^(1/2)}/(2x-6)

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shubhamsrg
  • shubhamsrg
well it should be -1/2 only cant be 1/2 as for x tending to -(a very large no.) ,,the expression gets negative,,clearly.. you can justify this as you may 4 = (-2)^2 and send 4 under sqrt sign with - aside ..
shubhamsrg
  • shubhamsrg
if you understand what am saying.
anonymous
  • anonymous
For positive infinity it is 1/2 for negative infinity it is -1/2
anonymous
  • anonymous
@jpsmarinho This is answer of your question..
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jpsmarinho
  • jpsmarinho
What you did in first step? I don't understand.
anonymous
  • anonymous
squaring both sides
anonymous
  • anonymous
|dw:1338756191105:dw|
anonymous
  • anonymous
@jpsmarinho Now remove the square y and get the radical over whole function .
jpsmarinho
  • jpsmarinho
Aah right!!
jpsmarinho
  • jpsmarinho
Other doubt: why I can not consider power the numerator and denominator without equal the expression to some variable?

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