Finding the limit of a quotient as x goes to minus infinity

[cmkluza]

Homework Statement

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

Homework Equations

N/A

The Attempt at a Solution

To solve this, I start off by dividing everything by $x^3$:
Numerator becomes $\frac{\sqrt{9x^6 - x}}{x^3} = \sqrt{\frac{9x^6 - x}{x^6}} = \sqrt{9 - \frac{1}{x^5}}$
Denominator becomes $\frac{x^3 + 9}{x^3} = 1 + \frac{9}{x^3}$
As $x$ approaches $-\infty$:
Numerator becomes $\sqrt{9 - 0} = \sqrt{9} = 3$
Denominator becomes $1 + 0 = 1$
So, the entire limit should evaluate to $\frac{\sqrt{9}}{1} = 3$. Yet this is not the case. What am I doing wrong?

 

[andrewkirk]

$\frac{\sqrt{9x^6 - x}}{x^3} = \sqrt{\frac{9x^6 - x}{x^6}}$

That is not correct if $x$ is negative. How should you adjust it for $x$ negative?

 

[epenguin]

Yet this is not the case. Why not? That's the only thing wrong I can see.

 

[cmkluza]

That is not correct if $x$ is negative. How should you adjust it for $x$ negative?

Ah, now I remember. Because $x$ is approaching minus infinity, it will be negative: $x^3$ should be negative if $x$ is negative, but $x^6$ would return a positive number. So, I just need to switch the sign in that equation, right? $\frac{\sqrt{9x^6 - x}}{x^3} = -\sqrt{\frac{9x^6 - x}{x^6}}$ Therefore the answer is -3, not 3.

I don't know why, but I always forget to consider positives/negatives when working around square roots. Thanks a bunch for your clarification!

 

[epenguin]

Hope you don't feel so alone now.

 

[Mark44]

To solve this, I start off by dividing everything by $x^3$:
Numerator becomes $\frac{\sqrt{9x^6 - x}}{x^3} = \sqrt{\frac{9x^6 - x}{x^6}} = \sqrt{9 - \frac{1}{x^5}}$
Denominator becomes $\frac{x^3 + 9}{x^3} = 1 + \frac{9}{x^3}$

$\frac{\sqrt{9x^6 - x}}{x^3} = \frac{|x^3|\sqrt{9 - \frac 1 {x^5}}}{x^3}$
The fraction $\frac{|x^3|}{x^3}$ evaluates to -1 if x < 0, and 1 if x > 0.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

N/A

The Attempt at a Solution

To solve this, I start off by dividing everything by $x^3$:
Numerator becomes $\frac{\sqrt{9x^6 - x}}{x^3} = \sqrt{\frac{9x^6 - x}{x^6}} = \sqrt{9 - \frac{1}{x^5}}$
Denominator becomes $\frac{x^3 + 9}{x^3} = 1 + \frac{9}{x^3}$
As $x$ approaches $-\infty$:
Numerator becomes $\sqrt{9 - 0} = \sqrt{9} = 3$
Denominator becomes $1 + 0 = 1$
So, the entire limit should evaluate to $\frac{\sqrt{9}}{1} = 3$. Yet this is not the case. What am I doing wrong?

Sometimes it helps to convert everything to positives, by putting $x = -t$ and taking $t \to +\infty$. That does not really change anything, but it eliminates one possible source of confusion/error.