This is a MATH137 (Calculus 1 for Honors Mathematics) quiz question from University of Waterloo, I did this quiz not too long ago and was going over the quiz today wondering why I got this wrong so I appreciate the video.
I like the explanation you had right at the end: “It’s too late, you can’t cancel anymore.” The value of a limit (assuming the limit exists) is just a number, and when you have one number divided by another number, we know from elementary school that we can’t be dividing by 0. But when you have a fraction inside of a single limit, you may have an indeterminate form of a limit like “0/0” (“ square quotes “), and you can do some algebra simplifying like in the video to cancel out the indeterminacy or use L’Hospital’s Rule if appropriate to be able to evaluate a single limit.
@@tyramoonshine3174 Alternatively, you can use a known series expansion for each function, where you inevitably find that the first terms disappear allowing h to be factored out and cancelled, then the higher powers can be taken to 0, leaving just a constant divided by another constant (neither of which is 0). Although, to be fair, that's exactly why L'Hôpital's Rule works.
Yes ... with most "laws", students seem to memorize only the formula itself of the law (... = ...) but not the condition(s) which should be checked first .... Not only about this one, but in many many other cases as well. But for sure, they cannot be blamed, as 99% of time with the examples they met, the learnt laws work just fine ... until they don't :)
@@monishrules6580 When is the sine law not applicable? (unless you're talking about a different sine law lol) I'm assuming you're talking about a/sinA = b/sinB = c/sinC right?
Math is many times like a chemistry lab, to understand what will happen in a specific situation, sometimes you need to set up the reagents, let the reaction happen and render you the result. What remains after is what is true.
The limit property in question cant always be used, as both limits must exist, and the denominator's limit must be both finite and nonzero. If both numerator and denominators limits are zero or both are infinite you have an indeterminate form.
In the numerator, how do we know that the limit operator encompasses both terms f(a+h) and -f(a) without parentheses? Is it just standard to assume everything that follows is part of what is being limited?
@@chri-k lim a + b is different from lim (a+b) just like how for example 3a + b is different from 3(a+b). what you meant is lim (a+b) is the same as lim a + lim b if you remove the parentheses, you have to give the limit to both of them
Personally, I'd call it sloppy notation. The mere fact that someone asks the question is a strong hint that adding parentheses would eliminate any ambiguity.
A limit never states the rate of convergence, so u can let h = 2m with m->0 on the top say, and h = 3n with n->0 on the bottom; h->0 in both cases, but at different rates, and the result isn’t f’(a) Of course, the h is a dummy variable and can be replaced with any letter for the two separate limits
I'd say B is not equivalent because it's undefined. It has no value, and only values can be equal to something else. The denominator being zero is one special case of this but it also applies if f or g is individually undefined at the point a. Also, the reason we can cancel h when the limits are still combined is because the expression inside the limit is explicitly never evaluated at the destination value, only arbitrarily close to that value. So you never divide by zero within the limit expression, only numbers arbitrarily close to zero. Outside of the limit, when the limit of h alone has already evaluated to zero, then you're talking about a real actual zero, not any numbers close to it, so you cannot divide.
I would prefer to say that the expression in B produces a result that is indeterminate, rather than undefined. That helps distinguish the cases where an expression in x has a value for some particular value of x that would be 0/0 (for example), and yet the limit of the expression as x approaches that particular value exists. That allows us to think about entities such as 0/0 as being able to take on any value, but we need more work to be able to evaluate it (e.g. taking a limit of the expression). Compare that with entities such as 1/0 which truly is undefined.
The thing to remember is that inside the limit, h is never zero. It only approaches zero. So dividing by h is not dividing by zero. But outside the limit, the limit's value (if it converges) is just a number. So dividing by the limit IS dividing by zero,
You can use L'Hopital's rule only in a limit, not on actual 0/0 case with the one with separate limits would be. Also is it or supposes that we can take detonates.
3:15 Note that $lim_{h\to 0} g(h)$ have to be nonzero is not the *only* necessary condition for the equality to hold. In fact, the left hand side limit may exist even when neither the numerator limit nor the denominator limit exists
Pausing at around ~15 seconds in, and my assumption is that because we have broken up the limit, we have to process it first before anything else, and therefore it will always be a zero over zero outcome, and no limit math is possible.
Something you could do to show that it causes problems distributing the limit like that without care is the following. When you get to the part where in the denominator it is the limit of h as h goes to 0, you could put in some arbitrary nonzero multiplier on h and preserve equality. For example, you could replace h with 2h in the denominator at that step because the limit of h and the limit of 2h both equal 0 as h approaches 0. But then you could use that same limit rule backwards to recombine the quotient of limits into a single limit of a quotient - only this time, there is now the extra factor of 2 in the denominator. You could factor the 1/2 out of the entire limit to show that the original derivative equals 1/2 of itself. And that could be done with other numbers, so you could make the derivative almost any number you want.
well maybe but that is a little picky in the technical terms. And maybe that is right when talking about foundations as this is a definition question. Still it feels like you forced a strict interpretation that logically does not need to be there. our base definitions also should be as simple as possible to avoid unnecessary complications.
I don't think it can give the right answer. It has the same result as lim_h→0( f(a+h) - f(a) ) / lim_h→0( nh ) where n has any value. Yet if you try to use L'Hôpital's on that the answer you get depends on the arbitrary value of n.
I have an interesting question: At what n will the x^n>x! Be satisfied. Or in form of equation x^n=x! I'm a bit dumb to do it myself, so I hope you'll see it.
There is no specific constant value of n where a power function is always equal to the factorial function for all x-values. Otherwise there'd be no purpose to the factorial or gamma function, as we could just use a power function in its place.
You can use Stirling's formula: ln(x!) = x.ln(x) - x approximately. Compare that with ln(x^n) = n.ln(x) and you can see that you're trying to find when n.ln(x) > x.ln(x) - x. That will happen when n > 1 - x/ln(x), i.e. most of the time, assuming n is a natural number.
that's a very interesting question. My guess with zero rigorousness is that f(x) -> N : x -> a at least under some circumstances implies ^f(x) -> a : x -> N ( where ^f is inverse of f ) This seems to work for 2^x and N = 0 anyway
If a function f(x) is continuous and bijective, then, if N is in its codomain (i.e. you can get N from that function), a = f⁻¹(N). If N isn't in the function's codomain, a = lim(y➝N) f⁻¹(y)
So, basically, this is entirely because mathematicians think that two separate limits with the same condition can't interact, even though they use two separate limits with the same condition when defining the rule which explains how to handle a division of two functions. This is absolute absurdity. B is correct; your prof, and this UA-camr, are silly. The only way this might work is if the limit on top only applies to the left portion and not the right portion as well. That's the only way I can see this as having any merit at all, but I can't prove that the limit designation applies that way. Looks silly regardless.
The reason B is incorrect is because the limit of h as h approaches 0 is not 0+ or 0- It's only a number near 0 inside the limit. But the limit itself is equal to 0, and you can never divide by zero. Do you understand now?
real analysis says that the limit of a composite function can be split into the limit of separate functions if and only if each separate limit exists and the expression as a whole is well defined in the reals. real analysis should be part of precalc to answer those problems. epsilon delta and infinitesmal are necessary prequisites
Behind the scenes: November 26, 2023
ua-cam.com/video/WsyKreSGgPU/v-deo.html
This is a MATH137 (Calculus 1 for Honors Mathematics) quiz question from University of Waterloo, I did this quiz not too long ago and was going over the quiz today wondering why I got this wrong so I appreciate the video.
As a Western student, I somehow knew that was a uWaterloo course code.
I like the explanation you had right at the end: “It’s too late, you can’t cancel anymore.” The value of a limit (assuming the limit exists) is just a number, and when you have one number divided by another number, we know from elementary school that we can’t be dividing by 0. But when you have a fraction inside of a single limit, you may have an indeterminate form of a limit like “0/0” (“ square quotes “), and you can do some algebra simplifying like in the video to cancel out the indeterminacy or use L’Hospital’s Rule if appropriate to be able to evaluate a single limit.
Thank you!! 👍
@@tyramoonshine3174 Alternatively, you can use a known series expansion for each function, where you inevitably find that the first terms disappear allowing h to be factored out and cancelled, then the higher powers can be taken to 0, leaving just a constant divided by another constant (neither of which is 0). Although, to be fair, that's exactly why L'Hôpital's Rule works.
Very simple, yet very well explained. Congrats 😃
Yes ... with most "laws", students seem to memorize only the formula itself of the law (... = ...) but not the condition(s) which should be checked first .... Not only about this one, but in many many other cases as well. But for sure, they cannot be blamed, as 99% of time with the examples they met, the learnt laws work just fine ... until they don't :)
Yeah... like the sine law
@@monishrules6580 When is the sine law not applicable? (unless you're talking about a different sine law lol) I'm assuming you're talking about a/sinA = b/sinB = c/sinC right?
@@theweirdwolf1877 i can't find it,maybe i was mistaken but i think it was this ua-cam.com/video/YR9pJGmiKso/v-deo.htmlsi=TUMebxiv9Sn80AxH
Math is many times like a chemistry lab, to understand what will happen in a specific situation, sometimes you need to set up the reagents, let the reaction happen and render you the result. What remains after is what is true.
The limit property in question cant always be used, as both limits must exist, and the denominator's limit must be both finite and nonzero. If both numerator and denominators limits are zero or both are infinite you have an indeterminate form.
In the numerator, how do we know that the limit operator encompasses both terms f(a+h) and -f(a) without parentheses? Is it just standard to assume everything that follows is part of what is being limited?
lim a + b
is usually assumed to mean lim (a+b),
unless that doesn't make sense
Not a very hard rule.
@@chri-k lim a + b is different from lim (a+b) just like how for example 3a + b is different from 3(a+b).
what you meant is lim (a+b) is the same as lim a + lim b
if you remove the parentheses, you have to give the limit to both of them
@@arceus1840 This depends on context ( since people are lazy and don't write parentheses ).
@@chri-k there is no laziness when it comes to science, especially when it comes to a quiz.
Personally, I'd call it sloppy notation. The mere fact that someone asks the question is a strong hint that adding parentheses would eliminate any ambiguity.
A limit never states the rate of convergence, so u can let h = 2m with m->0 on the top say, and h = 3n with n->0 on the bottom; h->0 in both cases, but at different rates, and the result isn’t f’(a)
Of course, the h is a dummy variable and can be replaced with any letter for the two separate limits
Yeah, choosing different limiting rates can actually end up with different answers in the end, which is why the limit B doesn’t exist or is undefined
My first thought is that probably in some cases the expression 'b' would lead to a 0/0 expression.
Hi, can you please tell me what app you used to write?
Good notes : )
Haven't finished watching but what if the function is non-differentiable at a...
the limit rules dont allow forming it like B.. if you bring down the term from the fraction you see that you have 2 limits. done.
Hum, isn't it obvious ? Lim h->0 h = 0
So we have a division by 0 that is not "protected" by any kind of limit, so it never exists.
I'd say B is not equivalent because it's undefined. It has no value, and only values can be equal to something else. The denominator being zero is one special case of this but it also applies if f or g is individually undefined at the point a.
Also, the reason we can cancel h when the limits are still combined is because the expression inside the limit is explicitly never evaluated at the destination value, only arbitrarily close to that value. So you never divide by zero within the limit expression, only numbers arbitrarily close to zero. Outside of the limit, when the limit of h alone has already evaluated to zero, then you're talking about a real actual zero, not any numbers close to it, so you cannot divide.
I would prefer to say that the expression in B produces a result that is indeterminate, rather than undefined. That helps distinguish the cases where an expression in x has a value for some particular value of x that would be 0/0 (for example), and yet the limit of the expression as x approaches that particular value exists. That allows us to think about entities such as 0/0 as being able to take on any value, but we need more work to be able to evaluate it (e.g. taking a limit of the expression). Compare that with entities such as 1/0 which truly is undefined.
Never disappoints.
I never liked multiple choice questions with more than one correct answer...
The thing to remember is that inside the limit, h is never zero. It only approaches zero. So dividing by h is not dividing by zero.
But outside the limit, the limit's value (if it converges) is just a number. So dividing by the limit IS dividing by zero,
very nice interpretation ty
Kind of funny that L’Hopital’s rule would work on the 0/0 because you’d get f’(a) from L’Hopital’s rule, which…yeah, sure hahaha
You can use L'Hopital's rule only in a limit, not on actual 0/0 case with the one with separate limits would be. Also is it or supposes that we can take detonates.
yeah sure what? what are you trying to communicate?
3:15 Note that $lim_{h\to 0} g(h)$ have to be nonzero is not the *only* necessary condition for the equality to hold. In fact, the left hand side limit may exist even when neither the numerator limit nor the denominator limit exists
Pausing at around ~15 seconds in, and my assumption is that because we have broken up the limit, we have to process it first before anything else, and therefore it will always be a zero over zero outcome, and no limit math is possible.
It is that simple.
Something you could do to show that it causes problems distributing the limit like that without care is the following.
When you get to the part where in the denominator it is the limit of h as h goes to 0, you could put in some arbitrary nonzero multiplier on h and preserve equality.
For example, you could replace h with 2h in the denominator at that step because the limit of h and the limit of 2h both equal 0 as h approaches 0.
But then you could use that same limit rule backwards to recombine the quotient of limits into a single limit of a quotient - only this time, there is now the extra factor of 2 in the denominator.
You could factor the 1/2 out of the entire limit to show that the original derivative equals 1/2 of itself. And that could be done with other numbers, so you could make the derivative almost any number you want.
That’s really a nice exercise, since it shows that someone underdtands limit laws
Agreed!
well maybe but that is a little picky in the technical terms. And maybe that is right when talking about foundations as this is a definition question. Still it feels like you forced a strict interpretation that logically does not need to be there. our base definitions also should be as simple as possible to avoid unnecessary complications.
Limits apply to functions, not parameters
So cool
i quickly figured out once i remembered the quotient rule for the limits and that important condition
0/0 is undefound
Funny it's not an equivalent yet still satisfies a L'Hopital's rule condition and gives the right answer.
I don't think it can give the right answer. It has the same result as lim_h→0( f(a+h) - f(a) ) / lim_h→0( nh ) where n has any value. Yet if you try to use L'Hôpital's on that the answer you get depends on the arbitrary value of n.
I have an interesting question:
At what n will the x^n>x! Be satisfied.
Or in form of equation x^n=x!
I'm a bit dumb to do it myself, so I hope you'll see it.
There is no specific constant value of n where a power function is always equal to the factorial function for all x-values. Otherwise there'd be no purpose to the factorial or gamma function, as we could just use a power function in its place.
You can use Stirling's formula: ln(x!) = x.ln(x) - x approximately. Compare that with ln(x^n) = n.ln(x) and you can see that you're trying to find when n.ln(x) > x.ln(x) - x. That will happen when n > 1 - x/ln(x), i.e. most of the time, assuming n is a natural number.
How would you tacle a problem like: lim x»a f(x) = N Find a
that's a very interesting question.
My guess with zero rigorousness is that
f(x) -> N : x -> a
at least under some circumstances implies
^f(x) -> a : x -> N
( where ^f is inverse of f )
This seems to work for 2^x and N = 0 anyway
If a function f(x) is continuous and bijective, then, if N is in its codomain (i.e. you can get N from that function), a = f⁻¹(N). If N isn't in the function's codomain, a = lim(y➝N) f⁻¹(y)
Doyou really spend 8 minutes to elaborate that the denominator is obviously zero?
So, basically, this is entirely because mathematicians think that two separate limits with the same condition can't interact, even though they use two separate limits with the same condition when defining the rule which explains how to handle a division of two functions. This is absolute absurdity. B is correct; your prof, and this UA-camr, are silly.
The only way this might work is if the limit on top only applies to the left portion and not the right portion as well. That's the only way I can see this as having any merit at all, but I can't prove that the limit designation applies that way. Looks silly regardless.
You don't understand limits.
@@kazedcat So helpful.
@@JDCullison1 I am pointing out facts. It is up to you to help yourself.
@@kazedcat *yawn*
The reason B is incorrect is because the limit of h as h approaches 0 is not 0+ or 0-
It's only a number near 0 inside the limit. But the limit itself is equal to 0, and you can never divide by zero. Do you understand now?
real analysis says that the limit of a composite function can be split into the limit of separate functions if and only if each separate limit exists and the expression as a whole is well defined in the reals. real analysis should be part of precalc to answer those problems. epsilon delta and infinitesmal are necessary prequisites