Question

I've just started Calculus this year and my teacher just taught a basic lesson on limits as x approaches infinity. In the lesson, she said that there are three rules that can be used to find the limit if the function is rational. The rules are as follows:
1) If the degree on top is bigger, the limit does not exist.
2) If the degree on bottom is bigger, the limit is 0.
3) If the degrees are the same, the limit is a ratio of the coefficients.
This is my question: Does that mean that with a function that has a square root in it, we can't use those rules to find the limit?

Answer 1

Of course you can. It's just a square root is a (1/2) power.

Answer 2

No @Psymon because if there is a square root of x in the algebraic function, then it cannot be a rational function!

Answer 3

See, that's what's got me tripped up. She went on a rant yesterday about how a rational function is a ratio of polynomials, and polynomials only have non-negative, integer exponents, which means no fraction exponents. ^What @calculusfunctions said. But it seems like I NEED to use it that way :(

Answer 4

I'm assuming he means that it's something like:

\[\frac{ x+2 }{ \sqrt{x-2} }\]

Answer 5

One sample question she gave us was \[(\sqrt{5x ^{2}+2x})/2x-7\]

Answer 6

Yeah, that would be fine. They show you to divide every term by the highest power in the denominator, right? Well....

Answer 7

@OddlyWeird A rational function is an algebraic function which is a ratio of two polynomial functions.

f(x) = a[n]xˆn + a[n-1]xˆ(n - 1) + a[n-2]xˆ(n - 2) + ... + a[1]x + a[0] is a polynomial function of a single variable, where a[0], a[1], ..., a[n-1], a[n] (a[0] are real valued coefficients (a[0] is also a constant term) and n belongs to the set of positive integers.

A rational function is a ratio of two polynomial functions. Hence
y = P(x)/Q(x) is a rational function if both P(x) and Q(x) are polynomial functions and Q(x) ≠ 0.

I). Factor, if possible, both P(x) and Q(x) completely.
II). State restrictions on the variables (values of x where Q(x) = 0).
III). Divide factors common to both P(x) and Q(x), to simplify.

If there are any factors in the denominator that completely disappear after dividing like factors, then the zeros of those factors indicate where there are removable discontinuities (I guess you can call them holes in the graph, although I don't like that).

If there are any factors that still remain in the denominator, the zeros of these factors indicate infinite discontinuities (vertical asymptotes). Graph of a rational function never intercepts a vertical asymptote. The function tends to either "+ve" or "-ve" ∞ as x "approaches" the vertical asymptote from either the left or the right.

If the degree of P(x) < the degree of Q(x), then there is a horizontal asymptote at y= 0 (the x-axis).
If the degree of P(x) = the degree of Q(x), then there is a horizontal asymptote at y= (leading coefficient of the numerator)/(leading coefficient of the denominator). In other words, the quotient of the leading coefficients.
If the degree of P(x) > the degree of Q(x), then a horizontal asymptote is no longer in existence, and makes way for an oblique asymptote. In other words, an oblique asymptote exists iff the degree of P(x) > the degree of Q(x). This should imply that a horizontal asymptote and an oblique asymptote never co-exist on the graph of a rational function.
the graph of a rational function is allowed to intercept a horizontal asymptote where the function equals the horizontal asymptote.

If the degree of P(x) is greater than the degree of Q(x) by 1, then we have an oblique asymptote which is on a linear diagonal run
(y = a[1]x + a[0]). You call this a slant asymptote (I don't like the word slant, either).
If the degree of P(x) is greater than the degree of Q(x) by 2, then we have an oblique asymptote which is a quadratic polynomial
(y = a[2]x² + a[1]x + a[0]).
If the degree of P(x) is greater than the degree of Q(x) by 3, then we have an oblique asymptote which is a cubic polynomial
(y = a[3]x³ + a[2]x² + a[1]x + a[0])...
In other words, oblique asymptotes are in fact, imaginary polynomial functions whose equations can be determined by long division. I suggest dividing the simplified function, for the ease of it. Also, like the horizontal asymptote, the graph of the rational function intercepts the oblique asymptote where the remainder is zero.

Answer 8

\[\frac{ \sqrt{5x ^{2}+2x} }{ 2x-7 }->\frac{ \sqrt{\frac{ 5x ^{2}+2x }{ x ^{2} }} }{ \frac{ 2x-7 }{ x } }\]
The reason I have an x^2 dividing in the numerator but an x on bottom is because in the bottom I am dividing by the highest power of x, which is 1, but in the top I am technically dividing by sqrt(x^2), which is the same as x. So I still could do this to check what my limit is if I'm careful.

\[\frac{ \sqrt{5+0} }{ 2-0 }->\frac{ \sqrt{5} }{ 2 }\]

Not trying to have my explanation interfere with @calculusfunctions, I just am trying to show how the same approach works because the degree on bottom is 1, but the numerator is just sqrt of the power of 2, which is the same as 1.

Answer 9

@OddlyWeird I just saw the sample question you posted above, and NO! That is NOT a rational function!

Answer 10

What @Psymon fails to understand is that the principle \[\sqrt{x ^{2}}=\left| x \right|\]which equals +x if x ≥ 0 and -x if x < 0.

Answer 11

I'm not out to argue about the definitions of it all, but what I did gives the correct answer :/ @calculusfunctions

Answer 12

@Psymon why do \[\frac{ 2 }{ x }\] and \[\frac{ 7 }{ x }\] equal zero?

Answer 13

@OddlyWeird
Because of the 3 things you mentioned in your original post. Once you have all of the separate terms, you take them all to infinity. And those all have a higher degree on bottom than what they do on top, so they go to 0.

Answer 14

@Psymon wait, what? I'm so sorry- I'm really slow at getting things. What is the degree of 7? I guess that's why I didn't know they'd equal zero. Also, you can take the individual limits of certain terms in the problem? I didn't know if that was okay or not

Answer 15

No @Psymon because the limit is -∞ if x → -∞, and the limit is +∞ if x → +∞, in the above square root function.

Answer 16

@OddlyWeird did you read my lesson above?

Answer 17

@Psymon are you a teacher?

Answer 18

Well, it's how I was shown to go about it. It's essentially the same thing you were told, just without the work involved. You divide every term in the problem by the highest degree in the numerator. After you have done so, anything with a higher degree remaining in the numerator goes to infinity, numbers just stay behind, and anything with a higher degree in the denominator goes to 0.

Answer 19

@calculusfunctions No, I am not, but I am just showing an approach to the correct answer. Is my answer wrong?

Answer 20

If
\[\frac{ \sqrt{5} }{ 2 }\]is wrong then please tell me how it is.

Answer 21

Yes but for non-rational functions, you must check the limit from both the positive and negative infinity.

Answer 22

Oh goodness. My head hurts O.o yes, I did read your lesson @calculusfunctions and thank you for typing all of that out! I'm just so confused about how to solve these problems. @Psymon has the right answer, but I don't understand how to do what you are saying to do

Answer 23

@calculusfunctions Okay, now we're getting somewhere. So even if the problem mentions limit to infinity, we still have to check negative infinity?

Answer 24

Well @Psymon I am a mathematician/Professor and yes your explanation is misleading, to say the least.

Answer 25

@calculusfunctions It would be great if you can explain your method and the correct approach....then we all can learn where psycon is making mistake

Answer 26

Alright @OddlyWeird post and example question here from your school notes, text, etc. and I will walk you through it.

Answer 27

@OddlyWeird And I'm sorry, I'm not here to argue with anyone or cause problems. I'm here to see what he has to say, I just know I've had a way of going about solving these problems that has worked for me and is what I remember being shown. If I'm doing something wrong then I'll see, but I just knew I had a way of trying to explain how to get the answer >.<

Answer 28

@Psymon don't be sorry! I completely understand. I appreciate your explanation! I'm just really confused about the entire topic, so I'm trying to find out at much as I can :)

Answer 29

Alrighty then, we have\[\lim_{x \rightarrow \infty}\sqrt[3]{\frac{ 4x ^{2}+1 }{ 3x ^{2}+4 }}\]Here we have an example of an odd root function. Thus limit as x -∞ can be treated the same as limit as x +∞. Hence @Psymon approach would work.\[\lim_{x \rightarrow \infty}\sqrt[3]{\frac{ 4x ^{2}+1 }{ 3x ^{2}+4 }}=\sqrt[3]{\lim_{x \rightarrow \infty}\frac{ 4x ^{2}+1 }{ 3x ^{2}+4 }}=\sqrt[3]{\frac{ 4 }{ 3 }}\]

Answer 30

But my teacher's answer sheet says \[\sqrt[3]{36}/3\], why?

Answer 31

However keep in mind that the the function you gave is still not a rational function.

Answer 32

Right, that is what you get after you rationalize the denominator.

Answer 33

Alright, so what's the approach if it were negative infinity?

Answer 34

Check the value of both answers on your calculator, and you should see that they are exactly the same! @OddlyWeird

Answer 35

@Psymon first of all, there is no need for you to ever apologize for trying to help. Secondly, to answer your question, for even roots, you should always check both, the limit from negative infinity and positive infinity.

Answer 36

So for the first question, going to negative infinity would result in the same answer but negative.

Answer 37

Yes @Psymon

Answer 38

Thank you both so much for all of your awesome help! I appreciated it SO, SO much. I'm probably going to end up posting another question soon because this whole worksheet threw me for a loop, so I'll be seeing you! :)

Answer 39

I think we got a slight bit more we can hear out first, though @OddlyWeird

Answer 40

Well guys, if that is all, I have to go now. Will return later when I have some time, so you can look for me then, if you need me. BYE!

Answer 41

Alright, later.

Answer 42

Can you solve \[\frac{ 2x-1 }{ \sqrt{4x^{2}+3} }\] the same way you solved the other equations?

Answer 43

To infinity or to negative infinity?

Answer 44

infinity

Answer 45

Alrighty. Well, my method involves dividing every term by the highest power in the denominator. So our highest power is going to be a square root of x^2, which as calc mentioned before, is going to be |x|. Now since we are going to infinity, x>0 and we can assume that this will be a positive x. Now I'll go about dividing every term by the highest power in the denominator:

\[\frac{ \frac{ 2x }{ x }-\frac{ 1 }{ x } }{ \sqrt{\frac{ 4x ^{2} }{ x ^{2} }+\frac{ 3 }{ x ^{2} }} }\]

Now I make sure to match the power of x to what it would be inside of the root and simplify it when outside of the root. And again, because we are going to infinity, we assumed that:
\[\sqrt{x ^{2}}=|x|=x\]We assumed positive. So now we just need to simplify what we have. After we simplify, we apply our limit as x goes to infinity.

\[\frac{ 2-\frac{ 1 }{ x } }{ \sqrt{4+\frac{ 3 }{ x ^{2} }} }\]
So 2 and 4 will just stay there since they dropped x terms. Now we take the other 2 terms to infinity. Because the degree of the denominator is higher in both of those fractions, as they go to infinity both will go to 0, leaving us with:
\[\frac{ 2-0 }{ \sqrt{4+0} }= \frac{ 2 }{ \sqrt{4} }=1 \]

Answer 46

WOW. I feel so stupid XD I wondered why I kept getting the wrong answer when I first tried to do it your way and it's because I divided the top by x^2 as well. Thank you for clearing that up!

Answer 47

Lol, it's fine. And it is the same thing you learned, which has the higher degree, numerator or denominator, it's just being careful that if you have sqrt(x^2), you can thinkk of the degree as being just 1 xD So yeah, be careful that you know whether or not you're dividing within the root or outside of the root.

Answer 48

What do you do if it's approaching negative infinity?