Question

what is the limit as x approaches negative infinity of (3x+sqrt(9x^2-x)), please no derivatives.

Answer 1

got it :)

Answer 2

\[\sqrt{9x^2 - x}+3x \times \left( \frac{ \sqrt{9x^2 - x} - 3x }{ \sqrt{9x^2 - x} -3x } \right)\]\[=\frac{ 9x^2 - x - 9x^2 }{\sqrt{9x^2 - x} - 3x } = \frac{ -x }{ \sqrt{9x^2 - x} - 3x } = \frac{ x }{ 3x -\sqrt{9x^2 - x}}\]\[= \frac{ x }{ 3x - x \sqrt{9 - \frac{ 1 }{ x }} } = \frac{ 1 }{ 3 -\sqrt{9 - \frac{ 1 }{ x }} }\]

i just realized that this only works if we take the negative root of sqrt(9) [3 - -3 = 6]

i'll look at it a bit more to figure it out. but as you can see it's 1/6

let me know if you have a question about any steps

Answer 3

thanks!