owlet
  • owlet
Question below:
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

owlet
  • owlet
1 Attachment
owlet
  • owlet
what's the next step?
owlet
  • owlet

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.