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- Thread starter Loki123
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How is 2) differ from 3)? The real question is which one do YOU think is incorrect and why?

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Take these two different limits one at a time. Since your main issue appears to be the signs in the answer, you need to pay close attention to signs everywhere.

Also, your markings don't necessarily explain your thinking fully. Words would help, as would taking smaller steps.

Here's an example of a specific error in what you wrote:

Presumably what you

just that in 2 we get x2 and in 3 we get 1/0. They could both be correct, after all we get the same result in the end, but i don't know that.How is 2) differ from 3)? The real question is which one do YOU think is incorrect and why?

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You cancelled x^2/x^2 in 3) why can't you do that in 2)??just that in 2 we get x2 and in 3 we get 1/0. They could both be correct, after all we get the same result in the end, but i don't know that.

right sorry, there has to be a lim after ever, except the last, equality sign. I forget that. I am asking this since this is how I am "supposed to" get the horizontal asymptote.The first question is, what limit are you trying to find? You are combining two different questions:

View attachment 30638

Take these two different limits one at a time. Since your main issue appears to be the signs in the answer, you need to pay close attention to signs everywhere.

Also, your markings don't necessarily explain your thinking fully. Words would help, as would taking smaller steps.

Here's an example of a specific error in what you wrote:

View attachment 30639

Presumably what youmeanis "[imath]\displaystyle=\lim_{x\to\pm\infty}x = \pm\infty[/imath]".

i got confused because we get x2 * 0, in 2.You cancelled x^2/x^2 in 3) why can't you do that in 2)??

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But you cancel the x^2 first as you did in 3) and you get 1/0. Am I misunderstanding?i got confused because we get x2 * 0, in 2.

Since the degree of the numerator is higher than the denominator. There's no horizontal asymptote, only slant/oblique asymptote.how I am "supposed to" get the horizontal asymptote.

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But all three methods show that there is no horizontal asymptote, which would require a finite limit.I am asking this since this is how I am "supposed to" get the horizontal asymptote.

The only difference between your results is whether the limits (in both directions) are both positive infinity, or one is positive and the other is negative. The latter is correct, as you can see by graphing the function:

The function approaches +infinity to the right, and -infinity to the left. Your writing, which combines both directions into one question, is inadequate for seeing things like this, which is why I said to separate it into two limits.

My point here was not just to complain about your not writing everything just right, but that if you take smaller steps and think about what you really mean when you cross things off, you can see better whether you are making a mistake. This will also communicate better to us, so we can tell what you are thinking.right sorry, there has to be a lim after ever, except the last, equality sign. I forget that.

In method 3, you have the denominator going to 0, but you didn't think at all, as far as I can tell, about whether it approaches zero

I think the good way to do it is oneWhich of these is correct (if any) and why? View attachment 30636

We were told to use separate cases only if we have log or e because it's the same for everything else.But all three methods show that there is no horizontal asymptote, which would require a finite limit.

The only difference between your results is whether the limits (in both directions) are both positive infinity, or one is positive and the other is negative. The latter is correct, as you can see by graphing the function:

The function approaches +infinity to the right, and -infinity to the left. Your writing, which combines both directions into one question, is inadequate for seeing things like this, which is why I said to separate it into two limits.

My point here was not just to complain about your not writing everything just right, but that if you take smaller steps and think about what you really mean when you cross things off, you can see better whether you are making a mistake. This will also communicate better to us, so we can tell what you are thinking.

In method 3, you have the denominator going to 0, but you didn't think at all, as far as I can tell, about whether it approaches zerofrom the positive or negative side. That is the answer to your question. That is why your third answer is wrong. The others are valid methods, though inadequately explained.

That is why I am confused. First is how I often do these problems and I get a nice answer, but my professor used the second approach and I am confused.0/0 is Not 1!

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10/2 = 5, since (only) 5*2 = 10

20/5 = 2, since (only) 2*5=10

9/0 has no answer, since 0*(no number) = 9

0/0 = 17, since 0*17=0.

0/0 = -2/3, since 0*(-2/3) = 0

0/0 = 1, since 0*1 = 0

...

We can't allow 0/0 to be any number. That is why we call it undetermined.

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I'd rather say, if the goal is to find aWe were told to use separate casesonly if we have log or ebecause it's the same for everything else.

You should be aware that

I don't like theThat is why I am confused.First is how I often dothese problems and I get a nice answer, butmy professor used the secondapproach and I am confused.

It really makes no sense, as you are keeping the x^2 in the numerator, but the one in the denominator disappears when multiplied by zero. Also, if you really had x^2, its limit would be positive in both directions. You need to explain your thinking, as I have asked.

Are you saying this is the professor's actual work, or just your own (wrong) idea of his method?

The

You factored out the highest power of x in each part, seeing that the rest approaches 0, and the powers of x simplify to x, which goes to positive x to the right and negative to the left. But, again, all that matters is that this is not finite, so there is no horizontal asymptote. You would not do this to find an oblique asymptote.

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I think it would be best to simply do the division or do to use l'hopital's rule.Which of these is correct (if any) and why? View attachment 30636

(x^2-4)/(x-1) = x+1 -3/(x-1). As x --> oo, 3/(x-1) approaches 0, 1 stays 1 and x goes to oo. Thus, the final answer would be oo

I think the good way to do it is one

sentences.

It's how it was written to the board. The student might have done it wrong without professor noticing. I don't know, I wasn't there at the time. Do you perhaps know of any videos (like khan academy or similar) when they take a function, find asimptotes, determine min and max y values and etc... I am really confused by this.I'd rather say, if the goal is to find ahorizontal asymptote, then in the case of a rational function, it will be the same in both directions, but for functions in general, it may not. So although that advice may be appropriate for your class, where you only see asymptotes in three cases, it is not good in general.

You should be aware thatlimitsare never stated with two different numbers being approached by x. Your problem is not really about limits at all, but about asymptotes, and you should have stated in that way from the start.

I don't like thesecondat all:

View attachment 30659

It really makes no sense, as you are keeping the x^2 in the numerator, but the one in the denominator disappears when multiplied by zero. Also, if you really had x^2, its limit would be positive in both directions. You need to explain your thinking, as I have asked.

Are you saying this is the professor's actual work, or just your own (wrong) idea of his method?

Thefirstis better:

View attachment 30660

You factored out the highest power of x in each part, seeing that the rest approaches 0, and the powers of x simplify to x, which goes to positive x to the right and negative to the left. But, again, all that matters is that this is not finite, so there is no horizontal asymptote. You would not do this to find an oblique asymptote.

It's how it was written to the board. The student might have done it wrong without professor noticing. I don't know, I wasn't there at the time. Do you perhaps know of any videos (like khan academy or similar) when they take a function, find asimptotes, determine min and max y values and etc... I am really confused by this.I'd rather say, if the goal is to find ahorizontal asymptote, then in the case of a rational function, it will be the same in both directions, but for functions in general, it may not. So although that advice may be appropriate for your class, where you only see asymptotes in three cases, it is not good in general.

You should be aware thatlimitsare never stated with two different numbers being approached by x. Your problem is not really about limits at all, but about asymptotes, and you should have stated in that way from the start.

I don't like thesecondat all:

View attachment 30659

It really makes no sense, as you are keeping the x^2 in the numerator, but the one in the denominator disappears when multiplied by zero. Also, if you really had x^2, its limit would be positive in both directions. You need to explain your thinking, as I have asked.

Are you saying this is the professor's actual work, or just your own (wrong) idea of his method?

Thefirstis better:

View attachment 30660

You factored out the highest power of x in each part, seeing that the rest approaches 0, and the powers of x simplify to x, which goes to positive x to the right and negative to the left. But, again, all that matters is that this is not finite, so there is no horizontal asymptote. You would not do this to find an oblique asymptote.

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Why use such method when you have l'Hospital?