# Thread: Calculus limits and negative infinity: lim[x->1](x-7)/(x2 + x - 2)

1. ## Calculus limits and negative infinity: lim[x->1](x-7)/(x2 + x - 2)

Hi,

sorry in advance if my layout below is basic/primitive, this is my first post.

I have been asked to "evaluate the following limit"

limit as x approaches 1 for (x-7)/(x2 + x - 2)

Substituting 1 for x I get -6/0 which is undefined so I tried to use the PSF method to factorise the denominator x2 +x - 2 which got me to (x+2)(x-1).

This didn't get me much further and still brings me to -6/0. The answer in my booklet tells me as x approaches 1 the function will equal negative infinity but I don't understand why this is.

Is it because as x approaches 1 the denominator approaches 0?

Thanks

2. #1 - Why did you substitute? This is a limit problem. There is no substitution until AFTER you prove continuity. Even then, I don't like it.

#2 - At some point in your mathematics career, you should have encountered the Rational Function. There is stuff to know from this pursuit.

#3 - PSF Method - Why does it matter HOW you factor? You should have factored in the first place. This would have told you that there are no common factors in the numerator and denominator. Thus, you are stuck with a vertical asymptote.

#4 - Why not try it out? What if x = 2? What if x = 1.5? What if x = 1.25? What if x = 1.125? and etc. Now, try the other side. x = 0? x = 1/2? x = 3/4? and etc. One fun thing to do is an experiment closer to the definition of the the unbounded limit. Pick a big number, say -1000. What values of x will produce this value? Okay, now try something bigger. -10,000? Is there a number so big that you CANNOT find a value for x to match it?

Study, survey, explore. Don't just memorize.

3. Well I am not trying to memorise, I am asking questions. I was pretty sure that is what this forum is for.

I probably did come across a Rational Function in the past but considering it has been over 20 years since I did high school maths the terminology, not to mention the concepts, have long since left my memory and I am now trying to pick things up again for university.

Am I out of my depth? Maybe but I thought asking a question would be a good place to start. I know I could attempt to answer the question by attempt what you suggest in #4 but I was wanted to know if there was a more sophisticated technique than that.

4. Originally Posted by veedubya
Hi,

sorry in advance if my layout below is basic/primitive, this is my first post.

I have been asked to "evaluate the following limit"

limit as x approaches 1 for (x-7)/(x2 + x - 2)

Substituting 1 for x I get -6/0 which is undefined so I tried to use the PSF method to factorise the denominator x2 +x - 2 which got me to (x+2)(x-1).

This didn't get me much further and still brings me to -6/0. The answer in my booklet tells me as x approaches 1 the function will equal negative infinity but I don't understand why this is.

Is it because as x approaches 1 the denominator approaches 0?

Thanks
Plot the function y = (x-7)/(x2 + x - 2) - say from -2 to +2

Tell us what you found.

5. Originally Posted by veedubya
Well I am not trying to memorise, I am asking questions. I was pretty sure that is what this forum is for.

I probably did come across a Rational Function in the past but considering it has been over 20 years since I did high school maths the terminology, not to mention the concepts, have long since left my memory and I am now trying to pick things up again for university.

Am I out of my depth? Maybe but I thought asking a question would be a good place to start. I know I could attempt to answer the question by attempt what you suggest in #4 but I was wanted to know if there was a more sophisticated technique than that.
Nothing wrong with plotting points. There are calculators and online resources to help you do this. Someday, sometime, when you are thinking about 4-dimensional objects, you will have to let go of the visual picture. It's fine for now.

It is hard work catching up to where you were. It can be frustrating - mostly because it doesn't happen fast enough. Patience and perseverance are required.

You do not appear to be out of your depth. Your judgment was good in thinking this a good problem for starters. It has already resulted in several useful conversations. Perfect.

6. ## negative infinity

Originally Posted by veedubya
Hi,

sorry in advance if my layout below is basic/primitive, this is my first post.

I have been asked to "evaluate the following limit"

limit as x approaches 1 for (x-7)/(x2 + x - 2)

Substituting 1 for x I get -6/0 which is undefined so I tried to use the PSF method to factorise the denominator x2 +x - 2 which got me to (x+2)(x-1).

This didn't get me much further and still brings me to -6/0. The answer in my booklet tells me as x approaches 1 the function will equal negative infinity but I don't understand why this is.

Is it because as x approaches 1 the denominator approaches 0?

Thanks
yes, and the numerator is negative.

7. Originally Posted by veedubya
Well I am not trying to memorise, I am asking questions. I was pretty sure that is what this forum is for.

I probably did come across a Rational Function in the past but considering it has been over 20 years since I did high school maths the terminology, not to mention the concepts, have long since left my memory and I am now trying to pick things up again for university.
Ah. See, that's kinda crucial information. People volunteer here to help students struggling with their homework. Thus, the usual assumption is that the poster is a student in a class, who has a basic grasp of the underlying and immediately preceding material. Clearly, we were mistaken in this case. (This is why we ask that people provide the context for their questions.)

When you say that you are "now trying to pick things up again for university", do you mean that you're trying to "study" for the placement test? Thank you!

8. Originally Posted by tkhunny
#1 - Why did you substitute? This is a limit problem. There is no substitution until AFTER you prove continuity. Even then, I don't like it.
I disagree. If the question were to give a formal proof, then you must formally prove continuity before you can substitute. There is, however, no obligation to prove everything from first principles when solving problems. When looking for the limit of a rational function, it is quite efficient to rely on the theorem that every polynomial is continuous and on the theorem that

$\displaystyle \lim_{x \longrightarrow a} f(x),\ \lim_{x \longrightarrow a} g(x) \in \mathbb R \text { and } \lim_{x \longrightarrow a} g(x) \ne 0 \implies \lim_{x \longrightarrow a} \left ( \dfrac{f(x)}{g(x)} \right ) = \dfrac{\displaystyle \lim_{x \longrightarrow a} f(x)}{\displaystyle \lim_{x \longrightarrow a} g(x)}.$

Using substitution to find that the limit of continuous g(x) is a non-zero real takes very little time and does not violate any rule of mathematics.

#4 - Why not try it out? What if x = 2? What if x = 1.5? What if x = 1.25? What if x = 1.125? and etc. Now, try the other side. x = 0? x = 1/2? x = 3/4? and etc. One fun thing to do is an experiment closer to the definition of the the unbounded limit. Pick a big number, say -1000. What values of x will produce this value? Okay, now try something bigger. -10,000? Is there a number so big that you CANNOT find a value for x to match it?
This I agree with enthusiastically. Numerical experimentation makes limits comprehensible. Moreover, numerical experimentation may suggest a limit and thus a way to demonstrate that limit.

9. Originally Posted by JeffM
Using substitution to find that the limit of continuous g(x) is a non-zero real takes very little time and does not violate any rule of mathematics.
No disagreement, but I would like a statement that there was even the tiniest thought about continuity prior to substitution. Substitution willy-nilly does violate fundamentals of mathematics.

10. Originally Posted by tkhunny
No disagreement, but I would like a statement that there was even the tiniest thought about continuity prior to substitution. Substitution willy-nilly does violate fundamentals of mathematics.
Well obviously I must agree with that. I was just focusing on the difference between calculating limits numerically with respect to functions that are everywhere continuous or are discontinuous only at obviously irrelevant points and formal proofs. It probably is important to stress to beginners that they should ponder a moment or two before assuming that they are dealing with a continuous function. And it would certainly be worthwhile to get students to focus on the fact that a function constructed from continuous functions may not be continuous itself.

The key point to me is that the limit of a ratio cannot be found from the limits of denominator and numerator unless both limits are real and the limit of the numerator is not zero. This is the fundamental difficulty that, for over 150 years, frustrated mathematicians trying to put calculus on a solid foundation.

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