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My dd is learning Calculus using Larson (w/Dana Mosley's instruction DVDs).

She is pretty good at figuring out where she made a mistake using the solutions manual. She can't figure out this one: 

lim (x®0) [ 1-cos x] /sin x = lim (x®0) [x/sin x] [1-cos x/x]  = (1)(0)= 0

 

 

She thinks it should be undefined : [0/0] [(1-1)/0] = unidefined

 

I can't always help her when she gets stuck as it has been 30 years since I last did calculus in college ?

Anyway, do you have a website where she could go to ask for help?

or if someone is good at this can you please let us know how this was solved.

Thank you!

 

 

 

Edited by StillStanding
adding parenthesis
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14 hours ago, StillStanding said:

lim (x®0) 1-cos x/sin x = lim (x®0) [x/sin x] [1-cos x/x]  = (1)(0)= 0

Are there supposed to be parentheses, i.e. is it really 1- cos x/sin x, or is it (1-cos x)/sin x?

I would use L'Hopital to solve this.

typo corrected

Edited by regentrude
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1 hour ago, regentrude said:

Are there supposed to be parentheses, i.e. is it really 1- cos x/sin x, or is it (1-cos x)/sin x?

I would use L'Hospital to solve this.

Yes there should be parenthesis ?

I will go back and correct it.

I checked the index in the book for "L'Hospital" and couldn't find the term. I will try to google it. 

How would you do it?

Thank you!

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4 hours ago, StillStanding said:

I checked the index in the book for "L'Hospital" and couldn't find the term. I will try to google it. 

 

L’Hopital page 2 of 11 http://staff.katyisd.org/sites/thscalculusap/Larson Chapter 8 Textbook/Ch 8.7 Indeterminate Forms and L'Hopital's Rule.pdf

ETA:

above link doesn’t work now. Click on ch 8.7 on this link http://staff.katyisd.org/sites/thscalculusap/Larson Chapter 8 Textbook/Forms/AllItems.aspx

Edited by Arcadia
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3 hours ago, Arcadia said:

The page doesn't exist when I click the link. I will check the Table of Contents in our book instead of the index. We also have the Steward book so I will look for it there too.

My daughter is in chapter 2 so there must be a different way to solve it...

Thank you for pointing us the right way!

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32 minutes ago, StillStanding said:

The page doesn't exist when I click the link. I will check the Table of Contents in our book instead of the index. 

 

If you can’t find it in your Larson book, go to this link http://staff.katyisd.org/sites/thscalculusap/Larson%20Chapter%208%20Textbook/Forms/AllItems.aspx and then click on Ch 8.7

DS13 also says to use the L’Hopital rule and he got 0 for the answer. 

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11 hours ago, StillStanding said:

Yes there should be parenthesis ?

I will go back and correct it.

I checked the index in the book for "L'Hospital" and couldn't find the term. I will try to google it. 

How would you do it?

Thank you!

sorry typo corrected. should be l-Hopital, no s. A calculus text that covers limits has to cover l-Hopital's rule. See here

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/lhopitaldirectory/LHopital.html

But I just saw in other post your DD is only in chapter 2. Then she would not be able to use l'Hopital because it requires derivatives. So without, you have to rearrange your expression until you can recover expressions for limits that you know because you had to calculate them in earlier problems. The solution you posted requires students to have found the limit x/sin x and cosx/x at some point prior and to know these are 1 and 0, respectively. 

These limits can be calculated in differed ways. You can make a simple geometric argument.

https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-8-limits-of-sine-and-cosine/MIT18_01SCF10_Ses8a.pdf

Or use use the squeeze theorem http://ime.math.arizona.edu/g-teams/Profiles/JS/Calc/SqueezeTheorem.pdf. I think one can also manipulate the exponential representation of the trig functions. 

Any of these was very likely done for the two functions at some point before this she encountered this problem.

Edited by regentrude
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Without knowing exactly what she has and has not learned yet, I can only guess as to how she's supposed to solve it, but the answer is indeed 0. [It looks like she tried to use the quotient rule without taking into account that she can't use it if the denominator equals 0.]

You can use the quotient rule, but only if the denominator doesn't equal 0...

lim (x->0) [(1 - cos x) / (sin x)]

lim (x->0) [((1 - cos x)/x) / (x/(sin x))]                        - multiply by x / x to prevent the denominator from being 0 later, and separate just like we did above

[lim (x->0) ((1 - cos x)/x)] / [lim (x->0) (x/(sin x))]     - apply the quotient rule

lim (x->0) ((1 - cos x)/x) = 0    and    lim (x->0) (x/(sin x)) = 0

0 / 1 = 0

Edited by aprilleigh
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2 hours ago, aprilleigh said:

Without knowing exactly what she has and has not learned yet, I can only guess as to how she's supposed to solve it, but the answer is indeed 0. [It looks like she tried to use the quotient rule without taking into account that she can't use it if the denominator equals 0.]

You can use the quotient rule, but only if the denominator doesn't equal 0...

lim (x->0) [(1 - cos x) / (sin x)]

lim (x->0) [((1 - cos x)/x) / (x/(sin x))]                        - multiply by x / x to prevent the denominator from being 0 later, and separate just like we did above

[lim (x->0) ((1 - cos x)/x)] / [lim (x->0) (x/(sin x))]     - apply the quotient rule

lim (x->0) ((1 - cos x)/x) = 0    and    lim (x->0) (x/(sin x)) = 0

0 / 1 = 0

She found you explanation very helpful! Thank you ?

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Sorry that I'm late, but this is actually not as bad if you use trig identities. 

The easiest way to do this is to recognize this as one of the formulas for tangent (x/2), which can be directly evaluated at zero. 

 If this doesn't pop into mind, you can also get there by multiplying by the conjugate of the numerator.

Multiply numerator and denominator by 1 + cos x. Numerator is now 1 - cos^2 x, which is equal to sin^2 x. Denominator is (1 + cos x)sin x. Cancel sin x from the numerator and denominator, and you now have sin x / (1 + cos x), and you can directly evaluate this at zero. 

This trick (conjugates) can often pop up in trigonometry problems -- if you see anything that looks like a fraction with 1 +/- sin x or 1 +/- cos x involved that is not immediately solvable by other means, see if this helps. 

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