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Mathematics
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infinite limit
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\[\lim_{x \rightarrow \infty}\sqrt{9x ^{2}-x}-3x\]
i want to say negative infinity but i cant prove it
Try to Remove the radical
i dont want a final answer, i want to know how to do it. so then its \[\left( 9x ^{2}-x \right)^{\frac{ 1 }{ 2 }}-3x\]
\[\sqrt{9x^2 - x}-3x * \frac{ \sqrt{9x^2 - x + }+3x }{ \sqrt{9x^2-x} +3x}\]
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how about \[\frac{\sqrt{9x^2-x}-3x}{1}\times\frac{\sqrt{9x^2-x}+3x}{\sqrt{9x^2-x}+3x}\]
Lol..@sirm3d
is that multiplying by the conjugate?
Yup
@Yahoo! you click faster than me LOL
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is it [\frac{ 9x ^{2}-4x }{ \sqrt{9x ^{2}-x}+3x }\]?
\[\frac{ 9x ^{2}-4x }{ \sqrt{9x ^{2}-x}+3x }\]?
i need an explanation, not the final answer
U should get |dw:1357560582212:dw|
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