The limit isn't infinity, that would have been the answer to the quiz and it would mean there is a limit, just not one in R. The limit doesn't exist in a much stronger sense than that, as the right limit isn't the same as the left limit.
This is mostly true. Sine is positive as it goes to zero from the right and negative as it goes to zero from the left, so the limit from the right of the entire thing is negative infinity and the limit from the left is positive infinity. This means that the (two sided) isn't either infinity and definitely does not exist at all. However, if the limit were infinity, many textbooks would still call that a limit that does not exist. In such books infinite limits are considered a type of non-existent limit instead of an existent limit. It makes the limit laws simpler to write if you define it that way because you can just say "if the limit exists..." instead of "if the limit exists and is finite... ." Of course, some textbooks take the approach that you did and say that the limit exists but it is not in the real line, and that is a valid way to write a textbook too. So, nothing you said was wrong, but one thing you said is author's preference instead of universal mathematics.
As x gets smaller, ln( 1 - x ) approaches -x. As x gets smaller, sin( x ) approaches x. So the numerator is approaching -2x. The denominator is basically sin( x )^2 which approaches x^2 as x gets smaller. So we get -2x / x^2 as x gets smaller. Getting smaller from the positive real side we get -2 / x, approaching -inf. Getting smaller from the negative real side we get 2 / abs( x ), approaching +inf. So limit does not exist.
It's easy to figure out the answer quickly because the denominator is sin^2x which like all trig functions is periodical and doesn't go anywhere. Therefore the limit can not exist
I like the way You hold Your Pen, So un-American :-) I was in gym class once and broke my ankle, I was messing about and couldn't tell. So I went through a whole school day limping. At one point I was shouted at and shed a tear to the laughter of all and I said sorry Sir. I stepped on some wood, it was a woodwork class. I was going down the drive 20 mins slower than the rest and our Gytm teacher came to me and asked why I was so tardy. Please Sir I hurt my leg being silly earlier. Wow! He took me to the hospital in his own car and I ended up at home with a plastered leg. We called it 'H' ospital. It was only later as a Maths geek, that I learnt it could be called 'Ospital, that's how I learned to kid folk that I knew French. You have a natural talent for explanation, you did a slick slight of hand with the ink and that was as neat as your blots were blots :-) ( ummm if that don't sound right .. it was meant to ..um sound right Imean).
3:30 You can't use L'Hopital's Rule on the limit as x goes to zero of sin(x)/x because that limit is the definition of the derivative of sin(x) at x=0. It is circular reasoning to use L'Hopital's rule for that limit because you are using the derivative to compute the limit, but since that limit is the derivative, so you have to first compute that exact limit to get the derivative in the first place.
It's only circular if the objective was to use l'hospitals rule to prove differentiability at zero. There's nothing wrong with using the rule to confirm the limit Edit: you're right, it does seem like it was being used as a proof
Why do you need to find limits. If you want to have a stable platform for a sphere the circular ring underneath is more stable. L hospital is if you take two sphere you need two rings or a single ring with the other side facing each other. Which means parallel lines.
Isnt it minus infinity??? This can be seen quickly if you remember that 1- cosx2 = sinx2, you can see then that the second part of the equation goes to minus infinity.
You can't use L'Hopital's Rule to determine the limit of sin(x)/x as x goes to zero because that limit is used to prove that the derivative of sin(x) is cos(x). This woman is using circular reasoning.
I don't see how her reasoning is circular. In the clip it is shown how sin(x)/x --> 1 as x-->0 follows from (1) L'Hopital's Rule and (2) sin'(x)=cos(x). (1) and (2) have proofs that do not depend on the limit of sin(x)/x. It would be circular to prove (2) by (1) in a naive fashion, but that is not what she is doing in the clip.
Didn’t understand a word. But you say it so beautifully I had to watch until the end.
Look, I’ve learned it, and I’ve forgotten it. I’ve learned it again. And I forgot it again. Plato said to Meno, “All learning is recollection.”
That's because Plato believed that there was an innate knowledge, hence we spend our lives recollecting it.
The limit isn't infinity, that would have been the answer to the quiz and it would mean there is a limit, just not one in R. The limit doesn't exist in a much stronger sense than that, as the right limit isn't the same as the left limit.
This is mostly true. Sine is positive as it goes to zero from the right and negative as it goes to zero from the left, so the limit from the right of the entire thing is negative infinity and the limit from the left is positive infinity. This means that the (two sided) isn't either infinity and definitely does not exist at all.
However, if the limit were infinity, many textbooks would still call that a limit that does not exist. In such books infinite limits are considered a type of non-existent limit instead of an existent limit. It makes the limit laws simpler to write if you define it that way because you can just say "if the limit exists..." instead of "if the limit exists and is finite... ."
Of course, some textbooks take the approach that you did and say that the limit exists but it is not in the real line, and that is a valid way to write a textbook too.
So, nothing you said was wrong, but one thing you said is author's preference instead of universal mathematics.
@@mathjazz6930 Verry well said Sir :-)
Unbelievably it came back to me 50 years after college.
i watched this entire video just because of the fountain pen and her handwriting.
lim x->0 = minus infinity. The first of the 3 final limits lim x -> 0 (x cosx - cosx -1) / 2 = (0 - 1 - 1)/2 = -1
Seems that when x-->0, x>0, the limit is -infinity and when x
As x gets smaller, ln( 1 - x ) approaches -x. As x gets smaller, sin( x ) approaches x. So the numerator is approaching -2x. The denominator is basically sin( x )^2 which approaches x^2 as x gets smaller. So we get -2x / x^2 as x gets smaller. Getting smaller from the positive real side we get -2 / x, approaching -inf. Getting smaller from the negative real side we get 2 / abs( x ), approaching +inf. So limit does not exist.
as x gets smaller ln(1-x) tends to ln(1) not to -x
without forgetting the condition that f and g are derivable on a and that both f(a)=g(a)=0
It's easy to figure out the answer quickly because the denominator is sin^2x which like all trig functions is periodical and doesn't go anywhere. Therefore the limit can not exist
I like the way You hold Your Pen, So un-American :-) I was in gym class once and broke my ankle, I was messing about and couldn't tell. So I went through a whole school day limping. At one point I was shouted at and shed a tear to the laughter of all and I said sorry Sir. I stepped on some wood, it was a woodwork class. I was going down the drive 20 mins slower than the rest and our Gytm teacher came to me and asked why I was so tardy. Please Sir I hurt my leg being silly earlier. Wow! He took me to the hospital in his own car and I ended up at home with a plastered leg. We called it 'H' ospital. It was only later as a Maths geek, that I learnt it could be called 'Ospital, that's how I learned to kid folk that I knew French. You have a natural talent for explanation, you did a slick slight of hand with the ink and that was as neat as your blots were blots :-) ( ummm if that don't sound right .. it was meant to ..um sound right Imean).
Dr. Fry?..what a treat you ar
What a pleasant person. You make a good TV math person, if there is such a thing. One question: do you real do math using a pen and not a pencil?
Terrence Howard the unqualified mathematician would poke holes in this
3:30 You can't use L'Hopital's Rule on the limit as x goes to zero of sin(x)/x because that limit is the definition of the derivative of sin(x) at x=0.
It is circular reasoning to use L'Hopital's rule for that limit because you are using the derivative to compute the limit, but since that limit is the derivative, so you have to first compute that exact limit to get the derivative in the first place.
It's only circular if the objective was to use l'hospitals rule to prove differentiability at zero. There's nothing wrong with using the rule to confirm the limit
Edit: you're right, it does seem like it was being used as a proof
We don't need LHospitals rule
We can use Taylor series expansion for ln(1-x) in t 1:46 he numerator and the whole expression would reduce to -2/x
See thenegative line on the feeding of a rular
Thank you,
Why do you need to find limits. If you want to have a stable platform for a sphere the circular ring underneath is more stable. L hospital is if you take two sphere you need two rings or a single ring with the other side facing each other. Which means parallel lines.
L'Hôpital says "both go to zero? No problem! Which goes FASTER?!"
Sin(x) is just x for small x
EXPAND IN A POWER SERIES!!!!
Isnt it minus infinity??? This can be seen quickly if you remember that 1- cosx2 = sinx2, you can see then that the second part of the equation goes to minus infinity.
-inf from the right side, +inf from the left.
1:10 LOL
What pens is she using?!
'are'?
S=9dfretheta
By induction, the rule can be used again so taking the second derivative of the numerator and denominator gives a limit of 0.5! 😀
Love a girl with a fountain pen.
1:13 CASEY??????????????????????????????????
Lol
You can't use L'Hopital's Rule to determine the limit of sin(x)/x as x goes to zero because that limit is used to prove that the derivative of sin(x) is cos(x). This woman is using circular reasoning.
I don't see how her reasoning is circular. In the clip it is shown how sin(x)/x --> 1 as x-->0 follows from (1) L'Hopital's Rule and (2) sin'(x)=cos(x). (1) and (2) have proofs that do not depend on the limit of sin(x)/x. It would be circular to prove (2) by (1) in a naive fashion, but that is not what she is doing in the clip.
HEAVY PAPER. FOUNTAIN PENS. MATHS.
Mathletes are hot
Nazarin was right & you were horribly, horribly, wrong.