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Can this problem be solved without calculus?


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My older kid says to use L'Hopital rule so I guess that falls under Calculus. His answer is 1/pi

 

That's what I'm getting too (L'Hopital's rule and 1/pi), but this was assigned by my son's precalculus teacher.

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If you know lim as u goes to 0 of sin(u)/(u) is 1, then you can solve it without l'hopital's rule. And usually that limit is taught in calculus (or even Precalculus) before derivatives are taught.

 

Yes--that limit has been taught.

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Then you mulitply the original limit by (pi x)/(pi x)

 

You should get lim as x goes to 0 of (sin(x)/x)((pi x)/sin(pi x))(1/pi)

 

Which is 1/pi

 

Where I get hung up is on the (pi x)/sin (pi x) piece.  Isn't sin (pi x) = 0 when x = 0?

 

(I should admit at this point that my trig is a bit...weak, or "developing" as they say in elementary ed.)

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Where I get hung up is on the (pi x)/sin (pi x) piece.  Isn't sin (pi x) = 0 when x = 0?

 

(I should admit at this point that my trig is a bit...weak, or "developing" as they say in elementary ed.)

x never gets to 0 that's the limit concept - try it in calculator with a very small x

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Where I get hung up is on the (pi x)/sin (pi x) piece. Isn't sin (pi x) = 0 when x = 0?

 

(I should admit at this point that my trig is a bit...weak, or "developing" as they say in elementary ed.)

Well yes. But the limit exists, even though the function is undefined. That's the concept of a limit. I tell my calculus students that the limit is what the function would want to be if the function was defined there. Since this is a hole in the graph instead of a jump discontinuity, the limit is defined.

 

Would you like it better if it was sin(pi x)/(pi x)? Since that limit is 1, then the limit of the reciprocal is also 1.

 

Are you looking for an epsilon-delta proof?

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Well yes. But the limit exists, even though the function is undefined. That's the concept of a limit. I tell my calculus students that the limit is what the function would want to be if the function was defined there. Since this is a hole in the graph instead of a jump discontinuity, the limit is defined.

 

Would you like it better if it was sin(pi x)/(pi x)? Since that limit is 1, then the limit of the reciprocal is also 1.

 

Are you looking for an epsilon-delta proof?

 

I'm looking for a way to do the problem using the knowledge that my son has (which is not at the level of an epsilon-delta proof--and, to be totally transparent here, my knowledge is also *well below* that level).

 

The way he has been taught to do these problems when the denominator is zero is to rejigger the expression to make it so when you plug in zero (in this case) for x, the denominator is not zero.  For example, you could show that (x^2-4)/x-2 when x goes to 2 would be 4 because when you cancel the (x-2)s and plug the 2 in, the result is 4.  

 

Is there something at the same level for this problem?  I'm obviously missing something--and it wouldn't be the first time!

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"Plugging in" the number is actually a bit of a trap in rational expressions.  It only works reliably when the variable can be isolated in either the numerator or denominator.

 

In this example, think about a multiplied function.  You should have seen lim( fg ) = lim( f ) x lim( g ), so you can break the complex problem into a product of known limits.

 

It's not an intuitively obvious problem if you've never seen the approach.  I would just chalk it up to "well, now we've seen a new technique - let's remember to consider it in the future."

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I'm looking for a way to do the problem using the knowledge that my son has (which is not at the level of an epsilon-delta proof--and, to be totally transparent here, my knowledge is also *well below* that level).

 

The way he has been taught to do these problems when the denominator is zero is to rejigger the expression to make it so when you plug in zero (in this case) for x, the denominator is not zero.  For example, you could show that (x^2-4)/x-2 when x goes to 2 would be 4 because when you cancel the (x-2)s and plug the 2 in, the result is 4.  

 

Is there something at the same level for this problem?  I'm obviously missing something--and it wouldn't be the first time!

 

If he hasn't learned that lim_{x -> 0} (sin x)/x = 1, it's not doable. You'll notice that that limit doesn't work by the technique you're mentioning either. You need to use that to do this without l'hospital's rule (and frankly, l'hospital's rule depends on the derivative of sine x which is usually derived with this limit, but that's neither here nor there). 

 

So here's what's inside your limit right now. You have (sin x)/(sin (pi x)). Rewrite this as (sin x / 1)(1/sin(pi x)) -- so two separate fractions. Multiply in pi x / pi x, which is 1, so this is legal. This can make your first fraction (sin x) / pi x and your second fraction (pi x / sin (pi x)). You can now split the limit into the product of lim_{x -> 0} (sin x) / pi x and lim_{x -> 0} pi x / sin (pi x). Let's consider these separately.

 

lim_{x -> 0} (sin x) / pi x = (1/pi) lim_{x -> 0} (sin x)/x = (1/pi) * 1 = 1/pi. 

 

For lim_{x -> 0} pi x / sin (pi x), let u = pi x. Then as x -> 0, u -> 0, so this limit is really lim_{u -> 0} u/sin u = lim_{u-> 0} 1/((sin u)/u) = 1/1 = 1. 

 

Multiplying these together means your original limit is 1/pi * 1/1. 

 

Here are some weblinks with more examples in better formatting. 

 

http://oregonstate.edu/instruct/mth251/cq/Stage4/Lesson/examplesSinx.html

 

http://patrickjmt.com/calculating-a-limit-involving-sinxx-as-x-approaches-zero/

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I followed everything up until the bolded:

 

For lim_{x -> 0} pi x / sin (pi x), let u = pi x. Then as x -> 0, u -> 0, so this limit is really lim_{u -> 0} u/sin u = lim_{u-> 0} 1/((sin u)/u) = 1/1 = 1. 

 

I don't understand why it's ok to take the reciprocal.

 

However, when I graph the two functions sinu/u and u/sinu, I see that while the functions themselves are not equivalent, the limit as u goes to 0 is the same.  

 

Wait--something that Caroline said upthread is suddenly making sense.  She said:

 

"Would you like it better if it was sin(pi x)/(pi x)? Since that limit is 1, then the limit of the reciprocal is also 1."

 

So, lim_{x->0} sinx/x and lim_{x->0} x/sinx both equal 1!  That makes sense!  (And is obvious, really, now that I'm thinking more clearly.)  

 

I want to thank you all for your explanations and for being so patient with me.   

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Curious - was this a problem from a Precalc book or is it some kind of challenge problem?

DS did some limits at the end of Pecalc but nothing as challenging as this problem.

 

Tell me about it.

 

It's an honors precalculus course, and they use a book that is maybe somewhat more challenging than a regular precalculus book, but not crazily so (the honors text is Demana and the regular text is Larson).  But the teacher only uses the problems in the book maybe 5% of the time.  The rest of the time it's worksheets that he makes up, gives no solutions for, with many problems going well beyond the instruction provided in the book or in class.  According to my son, most of the students in the class are having a lot of trouble (the teacher grades on a curve which makes it so parents can't really tell how their kids are doing unless their kids tell them).  The only reason my son isn't having trouble is because I do all of the problems ahead of him so that we have something resembling solutions or at least answers (so for this problem, if I hadn't gotten help here, I was still able to see using a graph what the answer was).  That way he is able to work to mastery rather than just getting an answer that may or may not be right. 

 

Sorry, that's probably way more than you ever wanted to know about my son's class.  It has been a major haul this year for reasons that have nothing to do with the material and everything to do with the teacher's ridiculous practices.

 

ETA:  And to answer the original question--it is a problem from a teacher-generated worksheet.

Edited by EKS
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I don't understand why it's ok to take the reciprocal.

 

Just to clarify, I wasn't taking the reciprocal really -- what I was doing was the equivalent of writing "2/3" as 1 over (3/2). To write it out with a little more space:

 

   u                    1

-------   =   --------------------

sin u                sin u

                       ------

                          u

 

Then since we know the numerator is a fixed number and the denominator tends to 1, the entire limit is 1/1 = 1.

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ETA: And to answer the original question--it is a problem from a teacher-generated worksheet.

When my math teachers did teacher generated worksheets, they wanted to prevent cheating from solutions manuals and they also indirectly wanted to see who has math help (tutors, parents, siblings and other relatives) after school. One of my 9th grade math teacher actually asked on the first day of class who has help at home.

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So glad you got it! I love when people get that aha moment in math.

 

I followed everything up until the bolded:

 

 

I don't understand why it's ok to take the reciprocal.

 

However, when I graph the two functions sinu/u and u/sinu, I see that while the functions themselves are not equivalent, the limit as u goes to 0 is the same.

 

Wait--something that Caroline said upthread is suddenly making sense. She said:

 

"Would you like it better if it was sin(pi x)/(pi x)? Since that limit is 1, then the limit of the reciprocal is also 1."

 

So, lim_{x->0} sinx/x and lim_{x->0} x/sinx both equal 1! That makes sense! (And is obvious, really, now that I'm thinking more clearly.)

 

I want to thank you all for your explanations and for being so patient with me.

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When my math teachers did teacher generated worksheets, they wanted to prevent cheating from solutions manuals and they also indirectly wanted to see who has math help (tutors, parents, siblings and other relatives) after school. One of my 9th grade math teacher actually asked on the first day of class who has help at home.

 

Yes.  I do think this is part of the issue.

 

But it doesn't make sense because he doesn't collect the homework--never sees it, has no clue who does well and who does not.  So if a student were to "cheat" by copying out of the solution manual, he'd get the same score on the homework as a student who got all of the answers wrong.  But both students would likely get a bad score on the tests.

 

The problem with the worksheets, aside from not having easily accessible solutions, is that it can be difficult to find a source of extra problems that are at the same level as the ones given by the teacher.  So when my son needs more practice with something that's non-standard (and this is usually what he needs extra practice with), he's out of luck.

 

My son would have flunked out of the course long ago if he didn't have my help at home.  It is frustrating because he is properly placed in the class--it's not the material, as I said before, but the lack of resources (solutions, extra problems, etc) that makes the course difficult.

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