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owlet
 one year ago
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owlet
 one year ago
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owlet
 one year ago
Best ResponseYou've already chosen the best response.0@SolomonZelman @Loser66 @freckles

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large\color{black}{\displaystyle\lim_{x \to~ \infty}\frac{\sqrt{2x^2+x+1}}{3x1}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1you can do this: \(\Large\color{black}{\displaystyle\lim_{x \to~ \infty}\frac{\sqrt{2x^2+x+1\color{white}{\LARGE }}}{3x1}}\) \(\Large\color{black}{\displaystyle\lim_{x \to~ \infty}\frac{\sqrt{2x^2+x+1\color{white}{\LARGE }}}{\sqrt{(3x1)^2}}}\) \(\Large\color{black}{\displaystyle\lim_{x \to~ \infty}\sqrt{\frac{2x^2+x+1}{(3x1)^2}}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large\color{black}{\displaystyle\lim_{x \to~ \infty}\sqrt{\frac{2x^2+x+1}{9x^26x+1}}=\sqrt{2/9}=\sqrt{2}/3}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1you can bring the limit inside the root, and do L'Hospital's rule twice.

owlet
 one year ago
Best ResponseYou've already chosen the best response.0how will i apply x approaches infinity though?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Ok, \(\large\color{black}{\displaystyle\lim_{x \to~ \infty}\sqrt{\frac{2x^2+x+1}{9x^26x+1}}=\sqrt{\lim_{x \to~ \infty}\frac{2x^2+x+1}{9x^26x+1}}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ \infty}\frac{2x^2+x+1}{9x^26x+1}}}\) You could apply L'Hospital's rule at this point, because as x\(\to \infty\), the top and bottom tend to ∞. (an ∞/∞ case)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1differentiate on top and bottom and you get \(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ \infty}\frac{4x+1}{18x6}}}\)

owlet
 one year ago
Best ResponseYou've already chosen the best response.0but we haven't learned l'hospital's rule yet

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Oh, so you aren't allowed to use it? (But do you know that rule, though?)

owlet
 one year ago
Best ResponseYou've already chosen the best response.0no, we're not allowed and I don't know that rule also

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ \infty}\frac{2x^2+x+1}{9x^26x+1}}}\) 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
 one year ago
Best ResponseYou've already chosen the best response.1It really did seem like an L'Hospital's rule kind of a problem, I apolgoize for mentioning it.

owlet
 one year ago
Best ResponseYou've already chosen the best response.0sorry 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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