# Thread: Dealing with modulus when trying to find limi

1. ## Dealing with modulus when trying to find limi

Apologies if this is posted in the wrong section.

I am trying to find the limit, as x approaches 3, of
$

(|5-2x|-|x-2|)/(|x-5|-|3x-7|)
$

Normally I would multiply/factorise/otherwise rearrange, but unfortunately I seem to have no clue how to add/multiply etc the modulus sections. Any suggestions would be much appreciated

Ivar

Ps: I'm not fussed about the actual answer (my calculator says it's -1/4) but interested in how to deal with the modulus parts

2. $x \in \left( {2.9,3.1} \right) \Rightarrow \quad \frac{{\left| {5 - 2x} \right| - \left| {x - 2} \right|}}
{{\left| {x - 5} \right| - \left| {3x - 7} \right|}} = \frac{{ - \left( {5 - 2x} \right) - \left( {x - 2} \right)}}
{{ - \left( {x - 5} \right) - \left( {3x - 7} \right)}}$

You should be able to finish.

3. Originally Posted by Ivar_sb
Apologies if this is posted in the wrong section.

I am trying to find the limit, as x approaches 3, of
$

(|5-2x|-|x-2|)/(|x-5|-|3x-7|)
$

Normally I would multiply/factorise/otherwise rearrange, but unfortunately I seem to have no clue how to add/multiply etc the modulus sections. Any suggestions would be much appreciated

Ivar

Ps: I'm not fussed about the actual answer (my calculator says it's -1/4) but interested in how to deal with the modulus parts
At x= 3, 5- 2x= 5- 6= -1 so 5- 2x is negative for x close to 3 and |5- 2x|= -(5- 2x)= 2x- 5.

At x= 3, x- 2= 3- 2= 1 so x- 2 is positive for x close to 3 and |x-2|= x-2.

At x= 3, x- 5= 3- 5= -2 so x- 5 is negtive for x close to 3 and |x- 5|= -(x-5)= 5- x.

At x= 3, 3x- 7= 9- 7= 2 so 3x- 7 is positive for x close to 3 and |3x-7|= 3x- 7

That means that
$\frac{|5-2x|-|x-2|}{|x-5|- |3x-7|}= \frac{2x-5 -(x-2)}{5- x- (3x-7)}$
$= \frac{x-3}{12- 4x}= \frac{x-3}{-4(x-3)}$
As long as x is not 3, we can cancel those "x-3" terms to get -1/4. The limit of that is, of course, -1/4.

4. ## Thanks

Thanks to both of you, it's actually quite simple once you see the trick. Doh.