Limit of function containing absolute values

gimme5minutes

Lim x->0 of (|4x-1| - |4x+1|) / x

How would I go about doing this, I cant even see how to start

gimme5minutes

I suppose I can break it up and make it

lim x->0 of |4x-1|/x - lim x->0 of |4x+1|/x

But what can I do then?

sponsoredwalk

Well the definition of a limit is that if both the left hand limit & the right
hand limit exist, are equal to one another then the limit exists.
Basically take them both seperately & see if they end up being equal.
If x < 0 then what happens to a function in absolute value signs,
how about if x > 0? Post your idea just in case as
you might not get it, I didn't straight away

edit: Where did the 2x - 1 etc... come from?

gimme5minutes

Sorry, misstyped it there. Well here's what I did -

(lim |4x-1| - lim |4x+1|) / lim x --- lim being lim x->0

So breaking it into approaching from left and right hand sides:

LHS
(lim |0-1| - lim |0+1|) / lim x, x->0-

|-1| - |1|) / lim x, x->0-

1 - 1 / lim x, x->0-

= 0/0

RHS
(lim |0-1| - lim |0+1|) / lim x, x->0+

|-1| - |1|) / lim x, x->0+

1 - 1 / lim x, x->0+

= 0/0

So the both LHS and RHS limits are 0/0, which doesnt exist. Is that correct? And is the way I calculated it correct?

sponsoredwalk

No that's not right, did you learn about indeterminate forms of the type [latex] \frac{0}{0} [/latex] ?

[latex] \lim_{x \to 0} \ \frac{|4x \ - \ 1| - \ |4x \ + \ 1|}{x}[/latex]

[latex] \lim_{x \to 0} \ \frac{|4x \ - \ 1|}{x} \ - \ \lim_{x \to 0} \ \frac{|4x \ + \ 1|}{x} [/latex]

If x < 0 then |4x - 1| < 0 & by definition

|4x - 1| = - (4x - 1) = 1 - 4x

That should be enough of a hint

gimme5minutes

sponsoredwalk wrote: »

No that's not right, did you learn about indeterminate forms of the type [latex] \frac{0}{0} [/latex] ?

[latex] \lim_{x \to 0} \ \frac{|4x \ - \ 1| - \ |4x \ + \ 1|}{x}[/latex]

[latex] \lim_{x \to 0} \ \frac{|4x \ - \ 1|}{x} \ - \ \lim_{x \to 0} \ \frac{|4x \ + \ 1|}{x} [/latex]

If x < 0 then |4x - 1| < 0 & by definition

|4x - 1| = - (4x - 1) = 1 - 4x

That should be enough of a hint

If x < 0 then |4x - 1| < 0 - do you mean 4x-1 < 0 as |4x-1| will always be positive? Im not sure I really get what your doing here. I dont know what the 1 - 4x you end up with signifies.

And I dont see how I can take what you've done and apply it to the right hand side limit, if I take x > 0 then |4x-1| > 0 is not necessarily true...for example, when x is less than 1/4 |4x-1| will be < 0.

sponsoredwalk

Ah, okay. Let me give you the definition of the absolute value of a function:

|x| = x if x > 0 & - x if x < 0.

This is the definition, you can also define it as:

|x| = √(x²)

both are worth knowing & helpful in different circumstances.

Basically the first definition is just saying that if x is greater than zero,
x > 0, then the number x is just x.

If x is less than zero, x < 0, then you take whatever negative number
that x is and make it negative (so that it becomes positive!). It's a
little confusing at first but you;ll get it.

You know the pythagorean theorem? a²+ b² = c²



well the line c can be expressed in terms of a & b:
c² = a²+ b²
c = √(a²+ b²)

This can also be written as:

|c| = √(a²+ b²)

Notice that this will always be a positive number! Why? Because the
square of any number, positive or negative, will become positive instantly &
then taking the square root of something positive is also positive!
The meaning of an absolute value is that you'll always have a
positive number or zero, but never ever a negative number. So, thinking about the definition I gave:
|x| = x if x > 0 & - x if x < 0.
We see that this is just sayng that no matter what number we have
we'll get a positive number taking the absolute value.
|3| = 3 because 3 > 0.
|-7| = - (-7) = 7 because (-7) < 0 so we take - (-7) = 7.
Notice that I can rewrite (-7) as (-3 - 4).
|-7| = |- 3 - 4| = - (- 3 - 4) = 3 + 4 = 7.

Now, in |4x - 1| if x < 0 we have to look at this and figure out what's
going on. Well if x is zero then we just get |-1| so it's will be negative.
If x is close to 0, say 1/16 away from 0, on the negative side, we have:
|4x - 1| = |4(-1/16) - 1| = |-(1/4) - 1| = |- (5/4)|
This is still negative. I think you can see that mentally anyway,
basically this will always be negative around zero on this side so you
know that |4x - 1| is always less than zero when approaching zero from
the left (going from the negative side up to zero!).
|4x - 1| < 0 means by definition we take - (4x - 1) = 1 - 4x.
Work with this value when finding the left hand limit,

[latex] \lim_{x \to 0^-} \ \frac{1 \ - \ 4x}{x} [/latex]

Re-read all of this if you have to, it will all just click sooner or later.
Look at this & tell me what happens when you approach zero from
the right, i.e. going from the positive side of the number line
down towards zero, will things change?

Also, look at |4x + 1|, how are things with that? Basically you work
with both fractions at a time together, what is the limit on the
left hand side of fraction A minus fraction B? Then what is the
limit from the right hand side of fraction A minus fraction B & if they
give the same number then it exists & is that number, whatever it is.
Remember, a function like (sinx)/x doesn't exist at x = 0 but it has
a limit at zero of 1

gimme5minutes

OK cool, I understand the |4x-1| < 0 part now and I think I get what Im supposed to be doing overall.

Work with this value when finding the left hand limit,

[latex] \lim_{x \to 0^-} \ \frac{1 \ - \ 4x}{x} [/latex]

1. Woud I be right in that as x tends towards 0 from the left, 1-4x = 1 and x (under the line) will be a very small negative number, thus giving this fraction a limit of - infinity?

2. looking at the limit on the left hand side of fraction B, |4x+1|/x, I cant say |4x+1| is positive or negative as it could be either...for example if x = -1 I'd get -3, whereas if x = -1/10 I'd get 6/10. So how do I deal with this?

sponsoredwalk

Well yeah the way I see it is that you are dividing by infinity for the first
fraction when x goes to zero. Unless I'm forgetting something there is no
way to deal with this so it's infinity.

As for the |4x + 1| you're right that when x = -1 you get |-3| and when x
= -(1/10) we get |4(-1/10) + 1| = |6/10| but the question is asking what
happens when x goes to zero so it's always safe just to try a few numbers
close to zero & see what happens. Like, (-1/100) is safe, but here so is
-1/10. Basically, when x is zero we get |1| so you'd need to have x lower
than -1 to affect the equation and we're interested in x around zero so it's
always positive. So 4x + 1 is what you'd work with & you'll see you get the
same nonsensical answer of infinity.

Still, the definition of a limit asks you to check both the left hand side
and the right hand side!!! So, as x approaches zero from the right
hand side means x as it goes from 1 down to 0.5 down to 0.25....
approaching zero, what is happning to the function?
If you get the same answer on both sides then that's the limit, if not
then nope!

gimme5minutes

OK thanks for all the help man, think I've got it worked out. I kept the fraction parts together and I ended up being able to cancel out the x ->

lim |4x-1| - |4x+1| / x

lim 1-4x - (4x+1) / x

lim -8x/x = lim -8 = -8

It worked out the same for the right hand limit, so my final answer is -8..correct?