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International Baccalaureatte | Maths HL
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Limits
Date Posted: 4 years ago
Views: 391
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You need to use the standard sine and cosine series to help you in the work.
All standard textbooks have this standard power series.
All standard textbooks have this standard power series.
Date Posted:
4 years ago
Allright thanks for the help!
The answer for ii) is incorrect. The limit should be 2.
Eric made the mistake in the series for sinx.
It should sinx = x - x³/3! + (x^5)/5! +... instead of sinx = 1 - x³/3! + (x^5)/5! ...
You will get 2 if done correctly.
Edit : this method cannot work if you don't divide both by x or use L'Hopital's rule on it as you will still end up with both the numerator and denominator being 0.
Alternative working using L'Hopital's rule directly :
lim ( (2x + 1)/sinx + 1/tanx )
= lim ( d/dx (2x + 1)/ d/dx(sinx) + d/dx(1) / d/dx (tanx) )
= lim ( 2/cosx + 0/sec²x)
= 2/cos0 + 0
= 2/1
= 2
Eric made the mistake in the series for sinx.
It should sinx = x - x³/3! + (x^5)/5! +... instead of sinx = 1 - x³/3! + (x^5)/5! ...
You will get 2 if done correctly.
Edit : this method cannot work if you don't divide both by x or use L'Hopital's rule on it as you will still end up with both the numerator and denominator being 0.
Alternative working using L'Hopital's rule directly :
lim ( (2x + 1)/sinx + 1/tanx )
= lim ( d/dx (2x + 1)/ d/dx(sinx) + d/dx(1) / d/dx (tanx) )
= lim ( 2/cosx + 0/sec²x)
= 2/cos0 + 0
= 2/1
= 2
Thanks J!
Just noticed another error in the answer for i) as well.
The terms crossed out should be (x - 1) but he has crossed out (x + 1) and (x - 1)
If crossed out correctly, you will get
limx→1 (x + 1)/(x - 2)
= (1 + 1)/(1 - 2)
= 2/(-½)
= -2
The terms crossed out should be (x - 1) but he has crossed out (x + 1) and (x - 1)
If crossed out correctly, you will get
limx→1 (x + 1)/(x - 2)
= (1 + 1)/(1 - 2)
= 2/(-½)
= -2
I got -2 initially and thought i was wrong. Thanks!!!
J : your alternative working for ( ii ) of 2/cos0 =2/1 =2 and not 0.
Yup, typo. Duly amended. The previous step was correct (2/1 which would give 2) so I'm sure Nonie would have noticed.
Welcome Nonie. Glad you got the answer correct on your own
Thanks Eric n J for your help!
Welcome.
Update :
Do take note that the series method doesn't work just by substitution alone as you will end up with both numerator and denominator as 0. And 0/0 is undefined.
So we have to either use L'Hopital's rule here, or do an extra step which is divide both numerator and denominator by x.
lim (x→0) [ (2x + 1 - cos x)/ sin x ]
= lim (x→0) [ (2x + 1 - (1 - x²/2! + x⁴/4! + ...))/(x - x³/3! + (x^5)/5! +... ) ]
From here it would lead to 0/0 if no division is done or if L'Hopital's rule isn't applied.
Dividing both numerator and denominator by x,
Then it becomes
lim (x→0) [ (2 + 1/x - (1/x - x/2! + x³/4! + ...) )/(1 - x²/3! + x⁴/5! +... ) ]
= lim (x→0) [ (2 + 1/x - (-x/2! + x³/4! + ...) )/(1 - x²/3! + x⁴/5! +... ) ]
= 2/1 (the other terms become 0)
= 2
Update :
Do take note that the series method doesn't work just by substitution alone as you will end up with both numerator and denominator as 0. And 0/0 is undefined.
So we have to either use L'Hopital's rule here, or do an extra step which is divide both numerator and denominator by x.
lim (x→0) [ (2x + 1 - cos x)/ sin x ]
= lim (x→0) [ (2x + 1 - (1 - x²/2! + x⁴/4! + ...))/(x - x³/3! + (x^5)/5! +... ) ]
From here it would lead to 0/0 if no division is done or if L'Hopital's rule isn't applied.
Dividing both numerator and denominator by x,
Then it becomes
lim (x→0) [ (2 + 1/x - (1/x - x/2! + x³/4! + ...) )/(1 - x²/3! + x⁴/5! +... ) ]
= lim (x→0) [ (2 + 1/x - (-x/2! + x³/4! + ...) )/(1 - x²/3! + x⁴/5! +... ) ]
= 2/1 (the other terms become 0)
= 2
J: Not true.
With the sinx as denominator, x from both num and denom cancelled out, leaving 2/1. Series is still applicable.
With the sinx as denominator, x from both num and denom cancelled out, leaving 2/1. Series is still applicable.
Yes, I meant to say it doesn't work without further manipulation.
In any study, there are always more than a single approach to solve the problem. Each method would require the person to fine tune his or her working to suit the question.
Most importantly, the student is taking exams so he/she should be using the methods that he/she can apply most intuitively/quickly and accurately to score under time constraint. Only exposure to different methods can allow one to decide which is better for him/her. This also increases her/his skill set and versatility.
As tutors/answer providers, the onus is on us to provide correct and sound working so that we don't end up teaching/imparting the wrong things.
As tutors/answer providers, the onus is on us to provide correct and sound working so that we don't end up teaching/imparting the wrong things.
While it is important that students need to get the correct answers in the shortest possible time, he or she must first understand and digest your single principle which he or she may not able to grasp. Every theory has it limitation. Thus an alternative in fact is prefer.
Perhaps you might not know, but L'Hopital's rule is taught in IB HL syllabus. So the student is expected to understand and know and use it.
J: a glance thro your list of unanswered questions accumulated over the months, maybe you can provide me a glimpse of your solutions of those unanswered.
Not sure what you mean by ' list of unanswered questions'.
I refer to the number of questions posted by student in your unanswered list!! Some questions were frozen for years. I believe those students have already graduated or promoted to the next stage of study
Is there such a thing as an unanswered list on this app?
Edit : interpreted your previous comment as personal unanswered list for each tutor/answer provider
Edit : interpreted your previous comment as personal unanswered list for each tutor/answer provider
I think you need to housekeep the unanswered list so that questions more than 6 months be dropped out from your accumulated list of questions.
I wonder are you the administrator of this program? If not, how is it that you are unaware of the "latest" and "unanswered" list??
Yes there is an unanswered list and latest list. But they aren't my lists. Lol.
You actually thought that I'm in charge of this app?
You actually thought that I'm in charge of this app?
Your access to our solutions and added comments and remarks pointed to the fact you are aware of the questions posted by the students and the comment by we answer providers. Maybe you would let me know your name so I can address you more respectfully. Are you working full time on this work - your instant replies on all my queries prompt me to think so?
Are you using manytutors through the mobile app , or the website?
I'm sure you realise as long as you registered an account with them, you can see all questions posted and all comments and answers uploaded?
You mean you aren't aware that the app has a notification function that alerts you when someone upvotes your answer / comments on your answer/ comments on the same question that you commented?
I'm sure you realise as long as you registered an account with them, you can see all questions posted and all comments and answers uploaded?
You mean you aren't aware that the app has a notification function that alerts you when someone upvotes your answer / comments on your answer/ comments on the same question that you commented?
You have not answermy question. How shd I address you?
Good morning J, I have yet to hear from you!
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