Limit of Absolute Values: Understanding the Concept

[azncocoluver]

Homework Statement

http://www4c.wolframalpha.com/Calculate/MSP/MSP621a026b7befdf1f1d00003b707661e1i20i4e?MSPStoreType=image/gif&s=55&w=126&h=38

Homework Equations

limit x->3- (3x-9)/((abs(3x-9))

The Attempt at a Solution

I don't understand how to find the limit of absolute values..

 

[lanedance]

what is the sign of |3x-9| for x<3?

 

[azncocoluver]

It would be -|3x-9| = -3x-9.. right?

 

[lanedance]

the sign is negative - however what you have written is in correct

teh way to do it is as follows
x<3
3x<9
3x-9<0

so it is negative, to make positive let's multiply by -1.

So for x<3
|3x-9| = -(3x-9) = 9-3x

 

[lanedance]

however an even simpler way would be to do a subtitution (maybe, depending what you're comfortable with...)

so you could let
y=3x-9

then
\lim_{x \to 3^-} \ \implies \ \lim_{y \to 0^-}

and the limit becomes
\lim_{x \to 3^-} \frac{3x-9}{|3x-9|} \ \implies \ \lim_{y \to 0^-} \frac{y}{|y|}

 

[azncocoluver]

the sign is negative - however what you have written is in correct

teh way to do it is as follows
x<3
3x<9
3x-9<0

so it is negative, to make positive let's multiply by -1.

So for x<3
|3x-9| = -(3x-9) = 9-3x

Okay so the fraction is now 3x -9 / 9-3x? How do I get the answer of -1 though? Also, is the limit of y/abs(y) is always going to be -1?

 

[lanedance]

1) how would you normally evaluate a limit, or where are you stuck. Is this one indeterminate?

2) No. Maybe from the negative side, but that's what we're trying to show isn't it.

Try out evaluating the direct limit first rather than the substitution, we can come back to that, but it is a helpful comparison

 

[azncocoluver]

Can you show me the steps to complete this problem because I have no idea where to start.

 

[lanedance]

well let's star my considering the numerator and denominator separately, qualitatively, what happens to each x goes to 3 from the left

the do you know about l'hopitals rule?

 

[azncocoluver]

We can't use L'Hopital's rule yet because my teacher hasn't taught it. Um, the numerator goes to infinity and so does the denominator. So infinity/infinity?

 

[lanedance]

not quite, have another think, what happens to 3x-9 when x->3

now this doesn't really need L'Hop as we have

\lim_{x \to 3^-} \frac{3x-9}{|3x-9|} = \lim_{x \to 3^-} - \frac{3x-9}{3x-9}= - (\lim_{x \to 3^-} \frac{3x-9}{3x-9})

as they're identical, i don't think it s a leap to say the final limit is -1

 

[azncocoluver]

Oh okay I think I finally understand. 3x-9 / 3x-9 is 1 and since there is a negative outside the parantheses, the limit is just -1 because you can't plug 3 into x since there is no x left.