Limit with Absolute Value

[hopelynnwelch]

I need to find the limit as x goes to 3/2+ of (2x+3)/|2x^2-3| and I suck at dealing with absolute values. So I watched some videos and most videos I find are like super simple...

My first instinct is to factor this:

(2x+3)/(|-x|*|2x-3|)

And even after watching videos I am still lost because even factoring and reversing signs nothing cancels.

Any tips on where to go with this?

 

[Ishuda]

I need to find the limit as x goes to 3/2+ of (2x+3)/|2x^2-3| and I suck at dealing with absolute values. So I watched some videos and most videos I find are like super simple...

My first instinct is to factor this:

(2x+3)/(|-x|*|2x-3|)

And even after watching videos I am still lost because even factoring and reversing signs nothing cancels.

Any tips on where to go with this?

What is |2x^2-3| when x is 3/2?

 

[hopelynnwelch]

Zero. But won't that make it undefined?

EDIT never mind... I looked at too fast and dropped the ^2

 

[hopelynnwelch]

Wait why 2x^2 -3

When I factor I get x(2x-3)

putting 3/2 in for that is zero...

 

[stapel]

Wait why 2x^2 -3

When I factor I get x(2x-3)

I see how one can factor an "x" out of "2x²", but how are you factoring an "x" out of the "3"?

 

[ClaireWu]

I need to find the limit as x goes to 3/2+ of (2x+3)/|2x^2-3| and I suck at dealing with absolute values. So I watched some videos and most videos I find are like super simple...

My first instinct is to factor this:

(2x+3)/(|-x|*|2x-3|)

And even after watching videos I am still lost because even factoring and reversing signs nothing cancels.

Any tips on where to go with this?

Just substitute x=3/2 into (2x+3)/abs{2(x^2)-3} gives (3+3)/[abs{2(9/4)-3}] = 6/[3/2] = 4 (Ans)

 

[Steven G]

Wait why 2x^2 -3

When I factor I get x(2x-3)

putting 3/2 in for that is zero...

x(2x-3) = x*2x - x*3 = 2x^2 - 3x which is not 2x^2 - 3

 

[hopelynnwelch]

I need to find the limit as x goes to 3/2+ of (2x+3)/|2x^2-3| and I suck at dealing with absolute values. So I watched some videos and most videos I find are like super simple...

My first instinct is to factor this:

(2x+3)/(|-x|*|2x-3|)

And even after watching videos I am still lost because even factoring and reversing signs nothing cancels.

Any tips on where to go with this?

EDIT I wrote the problem wrong... No wonder I'm so confused. Sorry

Should be:

(2x+3)/(2x^2-3x)

That's why I factored out an x from both terms.

When I do factor the x out and sub in 3/2 I do get zero.

 

[Steven G]

EDIT I wrote the problem wrong... No wonder I'm so confused. Sorry

Should be:

(2x+3)/(2x^2-3x)

That's why I factored out an x from both terms.

When I do factor the x out and sub in 3/2 I do get zero.

OK, so what is the limit?

 

[hopelynnwelch]

I factor this to

(2x+3)/|x|*|2x-3|

What I think I should do is look at the absolute values when x is greater or less than zero.

So x when greater than zero will just be x and x when less than zero will just be -x

Then I can do the same for 2x-3. If I set this equal to zero and solve I get 3/2.

So when x is greater than 2/3 it will be +(2x-3) when x is less than 2/3 it will be -(2x-3)

So then I have

(2x+3)/(x(2x+3)) and (2x+3)/(-x(-2x-3))

Does this look remotely right so far?

If it is when I sub in 3/2 I get zero meaning it is still undefined. This is where I am lost. When I change the signs there is no way to cancel or manipulate anything.

Hopefully I cleared that up a little I mistyped this like a billion times because the AV is throwing me off.

 

[Steven G]

I factor this to

(2x+3)/|x|*|2x-3|

What I think I should do is look at the absolute values when x is greater or less than zero.

So x when greater than zero will just be x and x when less than zero will just be -x

Then I can do the same for 2x-3. If I set this equal to zero and solve I get 3/2.

So when x is greater than 2/3 it will be +(2x-3) when x is less than 2/3 it will be -(2x-3)

So then I have

(2x+3)/(x(2x+3)) and (2x+3)/(-x(-2x-3))

Does this look remotely right so far?

If it is when I sub in 3/2 I get zero meaning it is still undefined. This is where I am lost. When I change the signs there is no way to cancel or manipulate anything.

Hopefully I cleared that up a little I mistyped this like a billion times because the AV is throwing me off.

You want to know what is going on when x is near (3/2)+ so why do you think you should be looking at x<0 and even x>0?

Can you write |2x-3| as a piecewise function? Once you answer this things will come together. If not with a little help it will.

 

[hopelynnwelch]

You want to know what is going on when x is near (3/2)+ so why do you think you should be looking at x<0 and even x>0?

Can you write |2x-3| as a piecewise function? Once you answer this things will come together. If not with a little help it will.

I don't know the absolute values are just confusing the heck out of me lol. Ive only done 50 of these lately and really thought I had it until I say an AV. Let me think about it as a piecewise and watch a few more videos and try it again.

Thanks Jomo

 

[Steven G]

I don't know the absolute values are just confusing the heck out of me lol. Ive only done 50 of these lately and really thought I had it until I say an AV. Let me think about it as a piecewise and watch a few more videos and try it again.

Thanks Jomo

Let * represent what is in between the absolute value bars

Then |*|= * if *>+0 or |*| = -* if *<0.

This means that what is in between the bars does not matter when it comes to getting the piecewise function.

For example |x+3| = x+3 if x+3>=0 or -(x+3) if x+3 <0

For example |7-y | = 7-y if 7-y >=0 or -(7-y) if 7-y <0

For example |2x+5|= 2x+5 if 2x+5 >=0 or -(2x+5) if 2x+5 <0

Notice that these 3 example are exactly the same except for what is in red and what is in red is always what is in between the absolute value bars.
Of course you have to clean things up, if necessary.

|x+3| = x+3 if x+3>=0 or -(x+3) if x+3 <0 becomes (solving for x) |x+3| = x+3 if x>=-3 or -x-3 if x<-3

For example |7-y | = 7-y if 7-y >=0 or -(7-y) if 7-y <0 becomes |7-y| =7-y if y<=7 or y-7 if y>7

For example |2x+5|= 2x+5 if 2x+5 >=0 or -(2x+5) if 2x+5 <0 becomes |2x+5| = 2x+5 if x>=-5/2 or -2x-5 if x<-5/2

Now try to write |2x-3| as a piecewise function.

 

[Ishuda]

I factor this to

(2x+3)/|x|*|2x-3|

...

I believe you are over thinking the problem. Assuming the proper form of the expression is
(2x+3)/|2x² - 3x|
we have that the denominator is always positive if x is not 3/2, otherwise it is zero. Thus the denominator goes to 0+ as x goes to 3/2 and, since the value of the numerator is positive at x = 3/2, the whole expression goes to positive infinity as x goes to 3/2.

 

[hopelynnwelch]

I believe you are over thinking the problem. Assuming the proper form of the expression is
(2x+3)/|2x² - 3x|
we have that the denominator is always positive if x is not 3/2, otherwise it is zero. Thus the denominator goes to 0+ as x goes to 3/2 and, since the value of the numerator is positive at x = 3/2, the whole expression goes to positive infinity as x goes to 3/2.

I went to my math tutor because this just drove me nuts...

And he pointed out that if X is undefined when I sub in 3/2 and there is no way to algebraically cancel anything, that I should just make a table.

You are right. I was way over thinking this so bad that I couldn't even remember to just sub in a value. When I sub 1.51, 1.501, and 1.5001 I get the limit is + infinity. For each of these the denominator is positive so I don't even have to change signs.

Thanks for all the help. Math always gets easier and easier and then they put bars around something and call it an ABS Value and just loose me... Sigh. Once again sorry for wasting your time on this one and writing the problem wrong a bunch of times. And thank you much for your time!

 

[Steven G]

I went to my math tutor because this just drove me nuts...

And he pointed out that if X is undefined when I sub in 3/2 and there is no way to algebraically cancel anything, that I should just make a table.

You are right. I was way over thinking this so bad that I couldn't even remember to just sub in a value. When I sub 1.51, 1.501, and 1.5001 I get the limit is + infinity. For each of these the denominator is positive so I don't even have to change signs.

Thanks for all the help. Math always gets easier and easier and then they put bars around something and call it an ABS Value and just loose me... Sigh. Once again sorry for wasting your time on this one and writing the problem wrong a bunch of times. And thank you much for your time!

Plugging in numbers is just to verify your results. Ishuda is absolutely right in post but I knew that you did not understand how to deal with absolute values that I wanted you work on that skill.
Here is the method using absolute values.
lim as x->3/2+ [(2x+3)/(|-x|*|2x-3|)]. Since we are concerned ONLY when x>3/2 we know that x>0 so |x| = x. Now 2x-3 >0 means 2x>3 which means x>3/2. What a coincidence--we want x>3/2. So |2x-3| =2x-3. Putting this all together we have
lim as x->3/2+ [(2x+3)/(|-x|*|2x-3|)] = lim as x->3/2+ [(2x+3)/(x*(2x-3))]= 6/0. Now we need to decide if denominator will be a little more than 0 or a little less 0. If x> 3/2 then x will be positive and 2x-3 will also be positive. The product will then be positive. So the lim is 6/0+ = positive infinity (positive since 6 is positive, 0+ is positive and +/+ is +)

 

[hopelynnwelch]

Plugging in numbers is just to verify your results. Ishuda is absolutely right in post but I knew that you did not understand how to deal with absolute values that I wanted you work on that skill.
Here is the method using absolute values.
lim as x->3/2+ [(2x+3)/(|-x|*|2x-3|)]. Since we are concerned ONLY when x>3/2 we know that x>0 so |x| = x. Now 2x-3 >0 means 2x>3 which means x>3/2. What a coincidence--we want x>3/2. So |2x-3| =2x-3. Putting this all together we have
lim as x->3/2+ [(2x+3)/(|-x|*|2x-3|)] = lim as x->3/2+ [(2x+3)/(x*(2x-3))]= 6/0. Now we need to decide if denominator will be a little more than 0 or a little less 0. If x> 3/2 then x will be positive and 2x-3 will also be positive. The product will then be positive. So the lim is 6/0+ = positive infinity (positive since 6 is positive, 0+ is positive and +/+ is +)

Yeah it has been five years since I have taken functions so I am rough all around with algebra and now jumping into Math 262. I am barely swimming by and can do most of these until I run into more complex versions (to my own head) and I just draw a blank. I did do well in functions with an A so I know I can do this but it is going to take a lot of work on my part as I am in no way math gifted and am now re-remembering much of this.

When I see you explain the logic of the problem I see how if I reason in this way I don't need values to arrive at a conclusion. I just need to understand how the function behaves when x is greater than 3/2 since both parts will be positive if x is greater than 3/2. Then all I have are polynomials and by this stage in the game I should be able to conclude that as x gets bigger the function can do one of two things, it can get bigger or smaller. In this case it gets bigger going to infinity. It takes me so long sometimes to reason simple things like this... Thanks for taking the time to help me understand it rather than just get through this one problem. I really appreciate that as well!

 

[Steven G]

Yeah it has been five years since I have taken functions so I am rough all around with algebra and now jumping into Math 262. I am barely swimming by and can do most of these until I run into more complex versions (to my own head) and I just draw a blank. I did do well in functions with an A so I know I can do this but it is going to take a lot of work on my part as I am in no way math gifted and am now re-remembering much of this.

When I see you explain the logic of the problem I see how if I reason in this way I don't need values to arrive at a conclusion. I just need to understand how the function behaves when x is greater than 3/2 since both parts will be positive if x is greater than 3/2. Then all I have are polynomials and by this stage in the game I should be able to conclude that as x gets bigger the function can do one of two things, it can get bigger or smaller. In this case it gets bigger going to infinity. It takes me so long sometimes to reason simple things like this... Thanks for taking the time to help me understand it rather than just get through this one problem. I really appreciate that as well!

Now work on some more!