# Finding a limit with a Square Root

$$\lim_{x\to \infty} \frac{\sqrt{9x^6-x}}{x^3+7}$$

I thought it would simply be $1/3$, not sure where I went wrong.

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To start: divide top and bottom by $x^3$. –  David Mitra Mar 17 at 17:51
I saw the limit at $-\infty$ before the edit. –  Sami Ben Romdhane Mar 17 at 17:58
@Sami Look at the review history; there was never any $-\infty$, but the OPs original post could have easily been misunderstood. –  amWhy Mar 17 at 18:03
@amWhy This's what the OP wrote: as lim -> -infinity (sqrt(9x^6-x))/(x^3+7) –  Sami Ben Romdhane Mar 17 at 18:05
Anyway thank you @amWhy I learned the use of the cancel command from one of you post and I tried use it in my answer here;-) –  Sami Ben Romdhane Mar 17 at 18:10

$$\lim_{x\to \infty}\frac{\sqrt{9x^6 - x}}{x^3 + 7}\quad = \quad\lim_{x\to \infty} \frac{\sqrt{9 - \frac{1}{x^5}}}{1 + \frac 7{x^3}} = \frac {\sqrt{9 - 0}}{1 + 0} = \sqrt 9 = 3$$
I divided the numerator and denominator by $x^3$.
Since $$x=_\infty o(x^6)\quad;\quad 7=_\infty o(x^3)$$ then we can write $$\require{cancel}\lim_{x\to-\infty}\frac{\sqrt{9x^6-x}}{x^3+7}=\lim_{x\to-\infty}\frac{\sqrt{9x^6\cancel{-x}}}{x^3\cancel{+7}}=\lim_{x\to-\infty}\frac{\overbrace{-3x^3}^{\ge0}}{x^3}=-3$$
The original post had $\;-\infty\;$ in the limit. This answer correctly addresses that. +1 –  DonAntonio Mar 17 at 20:17