Question

Limit question. Problem below.

Answer 1

what problem

Answer 2

He's posting it of course.

Answer 3

i know im just stupid

Answer 4

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

Answer 5

The answer i got was 3 and -3

Answer 6

\[
|x|=\sqrt{x^2}
\]

Answer 7

Oh now that you say it. Lemme redo the problem. I remember the teacher saying something about negative infinity then you just use negative.

Answer 8

Is it -3?

Answer 9

You can have a negative infinity limit that still is positive. Its just realizing that negative infinity would require you to use the rules of absolute value in this problem.

Answer 10

Since \(x\to -\infty\), we know \(|x|=-x\)

Answer 11

Here is how I did it:

Answer 12

So \(x = -|x|=-\sqrt{x^2}\)

Answer 13

\[
\lim_{x\to-\infty}-\sqrt{\frac{9x^2+2x}{x^2}}
\]

Answer 14

It goes to \(-3\), NOT \(3\)

Answer 15

\[\lim_{x \rightarrow - \infty}\frac{ -x \sqrt{9} }{ x }\]

Answer 16

-3

Answer 17

Thats what I just said. "Was it -3

Answer 18

I'm not going to just say the answer.

Answer 19

It's about the process.

Answer 20

But i just showed my work

Answer 21

Can't you see what i wrote

Answer 22

I don't see it.

Answer 23

\[\lim_{x \rightarrow - \infty}\frac{ \sqrt{x^2(9+\frac{ 2 }{ x }} }{ x }\]

Answer 24

\[\lim_{x \rightarrow - \infty}\frac{ -x \sqrt{9} }{ }\]

Answer 25

x got cut out on the bottom

Answer 26

and when you simplify ull get -3

Answer 27

That's true, but you don't pull out the \(x^2\)

Answer 28

... Ok, I understand the absolute values so I just went ahead and took it out and made it -.

Answer 29

So is this the answer?

Answer 30

\[\begin{align*}\lim_{x\to-\infty}\frac{\sqrt{9x^2+2x}}{x}&=\lim_{x\to-\infty}\frac{\sqrt{x^2}\sqrt{9+\dfrac{2}{x}}}{x}\\
&=\lim_{x\to-\infty}\frac{|x|\sqrt{9+\dfrac{2}{x}}}{x}
\end{align*}\]
As @wio mentioned, since \(x\to-\infty\), we know that \(|x|=-x\):
\[\begin{align*}\lim_{x\to-\infty}\frac{|x|\sqrt{9+\dfrac{2}{x}}}{x}&=\lim_{x\to-\infty}\frac{-x\sqrt{9+\dfrac{2}{x}}}{x}\\
\end{align*}\]

Answer 31

I see now. I was trying to pull the \(x\) in the denominator in, rather than trying to pull the \(x^2\) out. My bad. I didn't notice that route.

Answer 32

when x approaches inf ..you can take the squre of the highest degree of the radical and compare it with another variables in the equation