owlet
  • owlet
Question below:
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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owlet
  • owlet
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owlet
  • owlet
what's the next step?
owlet
  • owlet
@SolomonZelman @Loser66 @freckles

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SolomonZelman
  • SolomonZelman
\(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\frac{\sqrt{2x^2+x+1}}{3x-1}}\)
SolomonZelman
  • SolomonZelman
you can do this: \(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\frac{\sqrt{2x^2+x+1\color{white}{\LARGE |}}}{3x-1}}\) \(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\frac{\sqrt{2x^2+x+1\color{white}{\LARGE |}}}{\sqrt{(3x-1)^2}}}\) \(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\sqrt{\frac{2x^2+x+1}{(3x-1)^2}}}\)
SolomonZelman
  • SolomonZelman
\(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\sqrt{\frac{2x^2+x+1}{9x^2-6x+1}}=\sqrt{2/9}=\sqrt{2}/3}\)
SolomonZelman
  • SolomonZelman
you can bring the limit inside the root, and do L'Hospital's rule twice.
owlet
  • owlet
how will i apply x approaches -infinity though?
SolomonZelman
  • SolomonZelman
Ok, \(\large\color{black}{\displaystyle\lim_{x \to~ -\infty}\sqrt{\frac{2x^2+x+1}{9x^2-6x+1}}=\sqrt{\lim_{x \to~ -\infty}\frac{2x^2+x+1}{9x^2-6x+1}}}\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ -\infty}\frac{2x^2+x+1}{9x^2-6x+1}}}\) You could apply L'Hospital's rule at this point, because as x\(\to -\infty\), the top and bottom tend to ∞. (an ∞/∞ case)
SolomonZelman
  • SolomonZelman
differentiate on top and bottom and you get \(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ -\infty}\frac{4x+1}{18x-6}}}\)
owlet
  • owlet
but we haven't learned l'hospital's rule yet
SolomonZelman
  • SolomonZelman
Oh, so you aren't allowed to use it? (But do you know that rule, though?)
owlet
  • owlet
no, we're not allowed and I don't know that rule also
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ -\infty}\frac{2x^2+x+1}{9x^2-6x+1}}}\) You can just conclude the answer based on the leading coefficients, then. (The limit will be equal to 2/9 and the square root of 2/9 is going to be equal to √2 /3
SolomonZelman
  • SolomonZelman
It really did seem like an L'Hospital's rule kind of a problem, I apolgoize for mentioning it.
owlet
  • owlet
sorry i was off. I just finished my class. will it always work though? for those kind of problems? Just get the leading coefficients? I want to learn that rule.. where can I learn it? maybe I can just search it up on the internet.

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