zned6559
  • zned6559
Calculate lim t-> negative infinity ((2t^2 - t -2)^1/2) / (2t+1)
Mathematics
katieb
  • katieb
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amistre64
  • amistre64
x = sqrt(x^2) ... might be useful
zned6559
  • zned6559
I know the answer is -1/sqrt 2 but I don't know how to get there
amistre64
  • amistre64
^(1/2) is the same as sqrt() id put them both under the sqrt and compare them

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amistre64
  • amistre64
\[\frac{\sqrt{2t^2-t-1}}{\sqrt{(2t+1)^2}}\]
zned6559
  • zned6559
the bottom is suppose to be 2t+1 no sqrt
amistre64
  • amistre64
what does sqrt(x^2) reduce to?
zned6559
  • zned6559
x
amistre64
  • amistre64
then sqrt((2t+1)^2) = 2t+1
zned6559
  • zned6559
\[\sqrt{2t ^{2}-t-2}\div2t+1\]
amistre64
  • amistre64
you are not seeing the obvious ... \[\sqrt{2t ^{2}-t-2}\div2t+1\] \[\sqrt{2t ^{2}-t-2}\div \sqrt{(2t+1)^2}\]
amistre64
  • amistre64
what is the expansion of (2t+1)^2 ?
zned6559
  • zned6559
\[4t ^{2}+4t+1\]
amistre64
  • amistre64
then we have \[\sqrt{\frac{2t^2-t-2}{4t^2+4t+1}}\]
amistre64
  • amistre64
what do we know about polynomials and their limits?
zned6559
  • zned6559
not sure
amistre64
  • amistre64
then you need to cover some of your basic rules .... not sure what theyve covered tho
amistre64
  • amistre64
polys of the same degree, limit to their leading coeffs
zned6559
  • zned6559
I got it now!
amistre64
  • amistre64
the only thing to watch for is the sign, by squaring the bottom we altered the sign of it inthe process.
amistre64
  • amistre64
2t+1 as t to -inf is a negative value .... so we are originally going to have a negative limit in the end

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