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Mathematics 17 Online
OpenStudy (anonymous):

infinite limit

OpenStudy (anonymous):

\[\lim_{x \rightarrow \infty}\sqrt{9x ^{2}-x}-3x\]

OpenStudy (anonymous):

i want to say negative infinity but i cant prove it

OpenStudy (anonymous):

Try to Remove the radical

OpenStudy (anonymous):

i dont want a final answer, i want to know how to do it. so then its \[\left( 9x ^{2}-x \right)^{\frac{ 1 }{ 2 }}-3x\]

OpenStudy (anonymous):

\[\sqrt{9x^2 - x}-3x * \frac{ \sqrt{9x^2 - x + }+3x }{ \sqrt{9x^2-x} +3x}\]

OpenStudy (sirm3d):

how about \[\frac{\sqrt{9x^2-x}-3x}{1}\times\frac{\sqrt{9x^2-x}+3x}{\sqrt{9x^2-x}+3x}\]

OpenStudy (anonymous):

Lol..@sirm3d

OpenStudy (anonymous):

is that multiplying by the conjugate?

OpenStudy (anonymous):

Yup

OpenStudy (sirm3d):

@Yahoo! you click faster than me LOL

OpenStudy (anonymous):

is it [\frac{ 9x ^{2}-4x }{ \sqrt{9x ^{2}-x}+3x }\]?

OpenStudy (anonymous):

\[\frac{ 9x ^{2}-4x }{ \sqrt{9x ^{2}-x}+3x }\]?

OpenStudy (anonymous):

i need an explanation, not the final answer

OpenStudy (anonymous):

U should get |dw:1357560582212:dw|

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