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I don't understand why this limit is negative instead of positive

  • Thread starter mileena
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  • #1
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Homework Statement



lim [√(2x2 +1)]/(3x - 5)
x→-∞

I know that the answer is -√2/3. That is the answer my professor had on this example, and the math lab tutor also agreed, and so did my graphing calculator. But I keep getting √2/3 on paper. The math lab tutor explained yesterday why it is negative to me, but I didn't write it down, as I wanted to ask a few more questions, so I forgot why, which has left me irritated that I forgot.


Homework Equations





The Attempt at a Solution




lim [√(2x2 +1)]/(3x - 5)
x→-∞

lim √(2x2/x2 +1/x2)
x→-∞
/
(3x/x - 5/x)


lim √2 + 0
x→-∞
/
3 - 0


√2/3

The professor said something about the numerator being +∞ and the denominator being -∞, so the answer must be negative, but I thought the limit of the denominator was 3, which is positive?

Thanks for any help!
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



lim [√(2x2 +1)]/(3x - 5)
x→-∞

I know that the answer is -√2/3. That is the answer my professor had on this example, and the math lab tutor also agreed, and so did my graphing calculator. But I keep getting √2/3 on paper. The math lab tutor explained yesterday why it is negative to me, but I didn't write it down, as I wanted to ask a few more questions, so I forgot why, which has left me irritated that I forgot.


Homework Equations





The Attempt at a Solution




lim [√(2x2 +1)]/(3x - 5)
x→-∞

lim √(2x2/x2 +1/x2)
x→-∞
/
(3x/x - 5/x)


lim √2 + 0
x→-∞
/
3 - 0


√2/3

The professor said something about the numerator being +∞ and the denominator being -∞, so the answer must be negative, but I thought the limit of the denominator was 3, which is positive?

Thanks for any help!
Note that ##\sqrt{x^2} \neq x##!

In fact, ##\sqrt{x^2} = |x|##, so it equals ##x## if ##x \geq 0## and it equals ##-x## if ##x < 0##.

So your basic error occurred right at the start: you should have used
[tex]\frac{1}{x} \sqrt{2x^2+1} = -\sqrt{\frac{2x^2}{x^2} + \frac{1}{x^2}},[/tex] because ##x < 0##.

A much, much easier way is to note that for ##x \to -\infty## we have ##\sqrt{2x^2+1} \sim |x| \sqrt{2} = - \sqrt{2} x## and ##3x + 1 \sim 3x##. Basically, your professor had it right: the numerator is >0 and the denominator is < 0, so without thinking at all you KNOW the limit must be < 0.
 
  • #3
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Ok, thank you so much!

Basically, I think what you are saying is a square root can only return a positive value, so since the input to the square root is negative (-∞), which makes the entire inside negative in this case for some x's, if we want to avoid that undefined situation, we need to manipulate the situation to move the negative outside the radical.

So:

√52 = 5

but

√-52 ≠ -5

but

√-52 = -√52 = (-1)(5) = -5

But in the original problem, x is -∞. But since it is squared in the numerator, isn't it +∞, since (-∞)(-∞) = +∞ ?? So the limit should still be positive (as the denominator is still positive)?
 
  • #4
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0
A much, much easier way is to note that for ##x \to -\infty## we have ##\sqrt{2x^2+1} \sim |x| \sqrt{2} = - \sqrt{2} x## and ##3x + 1 \sim 3x##. Basically, your professor had it right: the numerator is >0 and the denominator is < 0, so without thinking at all you KNOW the limit must be < 0.
So, since √x2 = x, and √1 = 1, could I solve it this way:

Assuming this is true:

√(2x2 +1) =
(√2)(√x2) + √1 =
(√2)(x) + 1

I will do the following:

lim [√(2x2 +1)]/(3x - 5)
x→-∞



lim
x→-∞
-[[(√2)x]/x + 1/x]
/
3x/x - 5/x


lim
x→-∞
-[(√2) + 0]
/
3 - 0


lim
x→-∞
-[(√2)]
/
3


-√2/3

(Sorry, I can't figure out how to use LaTeX to make fancy fractions and radical signs like you did!)
 
Last edited:
  • #5
D H
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√-52 ≠ -5

but

√-52 = -√52 = (-1)(5) = -5
You just contradicted yourself. Taking it one step at a time, and using parentheses to avoid ambiguity, let's look at ##\sqrt{(-5)^2}##. ##(-5)^2=25##, and since ##\sqrt{25} = 5##, ##\sqrt{(-5)^2} = 5##.
 
  • #6
HallsofIvy
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You still seem to be very confused about basic definitions. First, you have
√-5^2 = -√5^2 = (-1)(5) = -5
No, that is NOT true. IF it were really "√-5^2= √-25" there would be NO value-a negative number does not have a (real) square root. But it is, in fact, √[(-5)^2]= √25= 5.

Then you say
But in the original problem, x is -∞.
.
No, it isn't. x cannot be -∞ because x is a number and "-∞" is not. The problem asks for the limit as x goes to -∞ which is just a short way or saying "x is negative with its absolute value getting larger with no bound".
 
  • #7
vela
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So, since √x2 = x, and √1 = 1, could I solve it this way:

Assuming this is true:

√(2x2 +1) =
(√2)(√x2) + √1 =
(√2)(x) + 1
That's not true. You really need to brush up on your basic algebra, otherwise it just makes learning what you're supposed to be learning that much harder.
 
  • #8
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0
Ok, thank you both for replying!

I seem to get more and more confused as the thread goes on, which speaks to my lack of mathematical ability.

It just seems to be in the original problem that all values used to compute the limit ended up being positive, so the answer should be positive as well.

In the numerator in the first term:

√2x2/x2

the x's canceled out, so all we were left with was √2

and for the second term in the numerator, we got √1/x2, which has a limit of 0. So √2 + √0 = √2, which is positive.

In the denominator, the x's also canceled out in the first term:

3x/x

so we got a limit of 3.

For the second term,

-5/x, the limit was 0.

so 3-0 = 3. Again, positive.

I am still trying to figure out where the negative in the final answer came from, using the method above I was taught to compute limits.
 
  • #9
D H
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The numerator is always positive. The denominator is negative for all x < 5/3, and therefore for all negative values of x. That means the ratio is negative for all negative values of x.
 
  • #10
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That's not true. You really need to brush up on your basic algebra, otherwise it just makes learning what you're supposed to be learning that much harder.
Oh, I know what I posted wasn't true, but I was trying to somehow agree with the first person that replied that I could divide both terms of the numerator by x instead of x2. So I needed a way to justify that, knowing that for limits, sometimes we have to manipulate things or suspend mathematical rules. Like for the this problem, where the x2 in the numerator was divided by x2, but the denominator was divided by x, which is mathematically not possible. But you had to do that to get the right result.
 
  • #11
vela
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Let's work backwards and start with
$$\sqrt{\frac{2x^2}{x^2}+\frac{1}{x^2}} = \sqrt{\frac{2x^2+1}{x^2}}.$$ When you pull the ##x^2## out of the square root, you don't get simply ##x##. You get its absolute value. This is Ray's point back in post 2. You end up with
$$\sqrt{\frac{2x^2+1}{x^2}} = \frac{\sqrt{2x^2+1}}{\lvert x\rvert}.$$ What you can't say is that
$$\sqrt{\frac{2x^2+1}{x^2}} = \frac{\sqrt{2x^2+1}}{x},$$ which is what you're doing. Because the limit is as ##x \to -\infty##, you have that ##\lvert x \rvert = -x##, so that
$$\sqrt{\frac{2x^2}{x^2}+\frac{1}{x^2}} = \frac{\sqrt{2x^2+1}}{\lvert x\rvert} = \frac{\sqrt{2x^2+1}}{-x}.$$ Does that make sense?
 
  • #12
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The numerator is always positive. The denominator is negative for all x < 5/3, and therefore for all negative values of x. That means the ratio is negative for all negative values of x.
Ok, I think I found the problem of what you are saying vs. the algorithm that I have been using.

For the numerator, I had

3x/x - 5/x =

3 - 0 =

3


But I also agree with what you said:

"The denominator is negative for all x < 5/3, and therefore for all negative values of x"

So the algorithm I was using (what was in my notes) was wrong I guess?
 
  • #13
vela
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Oh, I know what I posted wasn't true, but I was trying to somehow agree with the first person that replied that I could divide both terms of the numerator by x instead of x2. So I needed a way to justify that, knowing that for limits, sometimes we have to manipulate things or suspend mathematical rules. Like for the this problem, where the x2 in the numerator was divided by x2, but the denominator was divided by x, which is mathematically not possible. But you had to do that to get the right result.
This never happens. If you think you have to suspend a rule, it means you're making a mistake somewhere or misunderstanding something, and you need to figure out what that misconception is. Doing this will really help you sharpen your thinking and help you get rid of the notion that the math is mysterious and incomprehensible. It should always make logical sense to you.
 
  • #14
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Let's work backwards and start with
$$\sqrt{\frac{2x^2}{x^2}+\frac{1}{x^2}} = \sqrt{\frac{2x^2+1}{x^2}}.$$ When you pull the ##x^2## out of the square root, you don't get simply ##x##. You get its absolute value. This is Ray's point back in post 2. You end up with
$$\sqrt{\frac{2x^2+1}{x^2}} = \frac{\sqrt{2x^2+1}}{\lvert x\rvert}.$$ What you can't say is that
$$\sqrt{\frac{2x^2+1}{x^2}} = \frac{\sqrt{2x^2+1}}{x},$$ which is what you're doing. Because the limit is as ##x \to -\infty##, you have that ##\lvert x \rvert = -x##, so that
$$\sqrt{\frac{2x^2}{x^2}+\frac{1}{x^2}} = \frac{\sqrt{2x^2+1}}{\lvert x\rvert} = \frac{\sqrt{2x^2+1}}{-x}.$$ Does that make sense?
Thank you so much vela for taking the time to do that! I looked at it closely at first to make sure I understood everything. That made things much more clear for me!

I agree that √x = |x|. And that |x| = -x for negative numbers.

In this case though, am I right to say that for the numerator, in the end, it doesn't matter, as it is √2 - 0, which is positive (as the x's cancel out for the first term in the numerator and for the second term, the limit is 0)??

Is the reason the overall answer is negative is that the denominator is negative (although using the algorithm I have been taught, it is positive)?
 
  • #15
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This never happens. If you think you have to suspend a rule, it means you're making a mistake somewhere or misunderstanding something, and you need to figure out what that misconception is. Doing this will really help you sharpen your thinking and help you get rid of the notion that the math is mysterious and incomprehensible. It should always make logical sense to you.
Thank you for the encouragement!

I guess I was trying to figure out why the numerator had to be divided by x2, whereas the denominator had to be divided by just x. (Normally, you would divide both the numerator and denominator by the same term, so that term equals 1. Like if you divide both sides by x, you are really just multiplying by x-1/x-1 = 1. In my problem posted, we were actually multiplying by x-2/x-1 ≠ 1! But I think you're mathematical demonstration a few posts ago taught me how to convert that x-1 legally to x-2!
 
  • #16
vela
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No, you want to think about this a little more deeply. Keep in mind we're giving you two ways to see why the answer has to be negative. D H and Halls gave you the generally more useful and intuitive way of looking at it: when ##x## is large and negative, the denominator is negative while the numerator is positive, so the quotient is negative. You should, however, be able to show that the result is negative using the normal rules of algebra you're familiar with. The method you were learned in class is fine, but you have to be careful because of the square root.

You started with
$$\lim_{x \to -\infty} \frac{\sqrt{2x^2+1}}{3x-5} = \lim_{x \to -\infty}\frac{\frac{\sqrt{2x^2+1}}{x}}{\frac{3x-5}{x}}
= \lim_{x \to -\infty}\frac{\frac{\sqrt{2x^2+1}}{x}}{3-\frac{5}{x}}.$$ So far, so good. The problem arose in your next step when you basically said
$$\frac{\sqrt{2x^2+1}}{x} = \frac{\sqrt{2x^2+1}}{\sqrt{x^2}}.$$ Do you see why? What's the sign of the lefthand side? What's the sign of the righthand side?
 
  • #17
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No, you want to think about this a little more deeply. Keep in mind we're giving you two ways to see why the answer has to be negative. D H and Halls gave you the generally more useful and intuitive way of looking at it: when ##x## is large and negative, the denominator is negative while the numerator is positive, so the quotient is negative. You should, however, be able to show that the result is negative using the normal rules of algebra you're familiar with. The method you were learned in class is fine, but you have to be careful because of the square root.

You started with
$$\lim_{x \to -\infty} \frac{\sqrt{2x^2+1}}{3x-5} = \lim_{x \to -\infty}\frac{\frac{\sqrt{2x^2+1}}{x}}{\frac{3x-5}{x}}
= \lim_{x \to -\infty}\frac{\frac{\sqrt{2x^2+1}}{x}}{3-\frac{5}{x}}.$$ So far, so good. The problem arose in your next step when you basically said
$$\frac{\sqrt{2x^2+1}}{x} = \frac{\sqrt{2x^2+1}}{\sqrt{x^2}}.$$ Do you see why? What's the sign of the lefthand side? What's the sign of the righthand side?
I see that x = √x2 only if x is positive.

But if x is negative, then x ≠ √x2, since the square root function returns only positive values of x.

So, using the rules I have been taught, maybe the numerator is negative actually, because I am still getting a positive denominator using my algorithm, even though I understand that it is negative. So either the algorithm I have been using is wrong and the denominator is really negative; or my algorithm is correct but the numerator is negative.
 
  • #18
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Ok, I think that the numerator in my problem is actually negative:

√(2x2 + 1)
/
x


But since x = -∞:

√(2x2 + 1)
/
-|x|


√(2x2 + 1)
/
-√X2


-√(2 + 0)


-√2


So the overall answer is negative, as the denominator is positive (3 - 0 = 3).
 
  • #19
D H
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You missed a possibility: Your algorithm is wrong and the numerator is negative after dividing numerator and denominator by a negative value.

Your algorithm is wrong because you implicitly assumed ##x=\sqrt{x^2}##. That's okay only if you *know* that x is non-negative. It is not valid if you don't know the sign of x. What is valid is ##\operatorname{sgn}(x)\,x=\sqrt{x^2}##, where ##\operatorname{sgn}(x)## is -1 if x is negative, 0 if x is zero, +1 if x is positive.
 
  • #20
vela
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The original denominator, ##3x-5##, will be negative in the limit, but if you divide it by ##x## which is also negative, it'll become positive. Similarly, the original numerator is positive, but when you divide it by x, you flip its sign. So you originally have +/- = -, or after dividing both the numerator and denominator by ##x##, you have -/+ = -. The overall sign remains negative.
 
  • #21
vela
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Ok, I think that the numerator in my problem is actually negative:

√(2x2 + 1)
/
x


But since x = -∞:

√(2x2 + 1)
/
-|x|


√(2x2 + 1)
/
-√X2


-√(2 + 0)


-√2


So the overall answer is negative, as the denominator is positive (3 - 0 = 3).
Exactly!
 
  • #22
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The original denominator, ##3x-5##, will be negative in the limit, but if you divide it by ##x## which is also negative, it'll become positive. Similarly, the original numerator is positive, but when you divide it by x, you flip its sign. So you originally have +/- = -, or after dividing both the numerator and denominator by ##x##, you have -/+ = -. The overall sign remains negative.
Yes, thank you! I think that is what I did above. And I think that's where all the confusion arose in my understanding in this thread!
 
  • #23
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You missed a possibility: Your algorithm is wrong and the numerator is negative after dividing numerator and denominator by a negative value.

Your algorithm is wrong because you implicitly assumed ##x=\sqrt{x^2}##. That's okay only if you *know* that x is non-negative. It is not valid if you don't know the sign of x. What is valid is ##\operatorname{sgn}(x)\,x=\sqrt{x^2}##, where ##\operatorname{sgn}(x)## is -1 if x is negative, 0 if x is zero, +1 if x is positive.
Thank you. Yes, I did mess up. :approve: I know you are trying to help, but right now some things are still above my head often.

As for "sgn" above, I am not even aware of that operation. Hopefully I won't have to learn it for a while; I am so overwhelmed right now!
 
  • #24
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Sgn or signum function is defined as a function that gives 1 for all values of x in (0,inf), 0 when x=0 and -1 for (-inf,0).
inf is just me being lazy and not writing ∞ in LaTeX....
 
  • #25
D H
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In other words, it's 0 for x=0, |x|/x for non-zero x.
 

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