Question

Why is the answer to number ten not positive?
http://cdn.kutasoftware.com/Worksheets/Calc/01%20-%20Limits%20at%20Infinity.pdf

Answer 1

It is a calculus question

Answer 2

Numerator > 0
Denominator < 0

Answer 3

Ok, so what i did was I got rid of the plus one and plus two because they don't matter when x approaches negative infinity, then I simplified the numerator to x times root 2 over 4x, then I cancelled the x's out, leaving root 2 over 4. Where does the negative come from? Where did I mess up?

Answer 4

If what I said makes any sense

Answer 5

How did you address the square root?

Answer 6

What do you mean?

Answer 7

Well, you have a square root of a square... so the value is always positive.

Answer 8

There is no "cancelled .. out". From the ORIGINAL statement, the numerator MUST be positive and the denominator MUST be negative. You must understand this condition before you start adjusting the expression.

\(\sqrt{x^{2}} = |x|\) unless you know that x > 0

Answer 9

This is what I did: \[(\sqrt{2x ^{2}+1})/4x+2\]

Answer 10

Then I dropped the +1 and +2

Answer 11

I hope not. You should have tried \(\sqrt{2x^{2}+2}/(4x+2)\) -- The parentheses are NOT optional.

Answer 12

Then I got \[(\sqrt{2x ^{2}})/4x\]

Answer 13

Here, take a look at a graph of it:

Answer 14

You are about to make the error.

Answer 15

What parentheses?

Answer 16

The parentheses in the denominator that you failed to write when you wrote the entire expression differently.

Answer 17

I dont follow

Answer 18

I didn't do anything to the denominator except drop the +2

Answer 19

I only put in parentheses to show it is a rational expression

Answer 20

\(\dfrac{\sqrt{2x^{2}+1}}{4x+2}\ne (\sqrt{2x^{2}+1})/4x+2\)

Answer 21

You just wrote it incorrectly.

Let's get back to \(\sqrt{2x^{2}}/4x\)

You are about to make the fatal error.

Answer 22

Ignore the parentheses, those aren't in the actual problem. I only put them there when I was typing because I didn't know how to make a fraction in the equation thing

Answer 23

Ok, so the "fatal error"

Answer 24

You are not seeing the incorrectness of the expression. The way you write it was INCORRECT. It is not a matter of interpretation.

What was your next step? That was the fatal error.

Answer 25

Ok, so I broke SQRT(2x^2) into SQRT(2) times SQRT (x^2)

Answer 26

And then I simplified it to xSQRT(2)

Answer 27

There is it! \(\sqrt{x^{2}} = x\) ONLY IF we know \(x>0\).

In this case, we KNOW x < 0, therefore \(\sqrt{x^{2}} = -x\)

Answer 28

Ehhhhhh ok

Answer 29

I figured I did something stupid

Answer 30

Ok, that makes MUCH more sense

Answer 31

So if x was approaching positive infinity, it would be a POSITIVE solution?

Answer 32

This symbol, \(\sqrt{}\) means the principle (or positive) square root. You had to know that up front or you were very likely to trip over it along the way.

Don't beat yourself up. Other may get this problem correct, but I suspect they will not know why. Be very careful with the ORIGINAL structure. Don't lose it along the way.

Positive leads to positive. That is right.

Answer 33

One definition of absolute value is:
\(|x|=\sqrt{x^2}\)

The logic being that the squaring makes it always positive.

That is what makes the graph look like it does.

Answer 34

Ok, I forgot about the plus or minus sq root thing. Thanks a lot for the help!!