You made a mistake I believe with the second use of L'hospital's rule. You differentiated xcosx to x*-sinx + sinx * 1 when it should be x*-sinx + cosx * 1. So the final answer should be 0/2
@@Rozenkrantzz Err, my example _has_ the same condition. x + x² ~ x as x -> 0. You can easily calculate that for yourself: (x+ x²)/x = 1 + x, and for x -> 0, that goes to 1.
@bjornfeuerbacher5514 they're kinda not the same, although this might be a little bit handwavy. x and sin(x) get arbitrarily close together as x approaches zero. But with x + x² that's kinda not the same, there's always this lingering small tail of x² that would make it different from 1/x at its limit. Or maybe not. But in any case, you're right, just saying that f(x) = g(x) at zero is not enough generally
@@arseniix For sin(x), there _also_ is such a "lingering small tail". If you look at the Taylor series of sin(x), that tail is x³/6. The point is that for sin(x), there is no terms with x². _That_ is why the limit works for sin(x), but not for x+x². And that's why your "proof" above is not enough, you have to invest more work.
For the sake of brevity, let lim f(x) denote the limit of f(x) as x→0. lim(csc x - 1/x) = lim ((x-sinx)/x³)(x/sinx)x. Since the limit of each of these three terms exists, by algebraic limit theorem, = [lim(x-sinx)/x³][lim(x/sinx)][lim x] = ⅙(1)(0) = 0
This may be sloppy, but we know lim x-> 0 sin(x)/x = 1. So the two terms in the original are essentially equal near 0. So their difference is essentially 0 near 0.
I see many comments mention the use of sin(x) ≈ x. Based on the video it intuitively cannot work because he used L'Hospital *twice* i.e. he differentiated twice, while sin(x) ~ x is a 1st order approximation (this is only an intuition, I never said it's a proof lol). However if we consider the next term in the Taylor series, sin(x) ≈ x - x³/6. Thus we get x/sin(x) ≈ x/(x - x³/6) = 1/(1- x²/6) ≈ 1 + x²/6 (In the last equality, use 1/(1-u) ≈ 1+ u) In the end we get 1/sin(x) - 1/x ≈ 1/x + x/6 - 1/x = x/6 -> 0.
Let f and g be 2 functions such that f(x) ~g(x) when x->0. Does it means that 1/f - 1/g - >0 when x->0? No. Here it does but only by chance. Set f(x) =x and g(x) =x+x^p for some p>1. f~g as x->0, ok. 1/f-1/g = 1/x * (1-1/(1+x^{p-1})). But 1/(1+u) ~ 1-u when u->0 then 1/f-1/g ~ x^{p-2} (don't forget the 1/x) which goes to 0 if p>2, 1 if p=2, and even infty for 1
You can do it without L'Hopital too ( Just for extra ) multiply divide by x+sinx in Nr and Dr then make the equation as x^2 -sin^2x / xsinx ( x+Sinx) rewrite it as x^2 - ( 1+cosx)(1-cosx)/xsinx(x+sinx) Now take x^2 common from the Nr and x from xsinx and another x from x+sinx You'll get it done
@@robertveith6383 umm sorry sir i didn't get what you mean , it would be great if you tell exactly in what line , although i 've tried it on notebook that's how i told it up here , maybe the grouping of numerator and denominator on whole remains wrong here yeah maybe but you can definitely get what i mean ( although thanks alot ) Grouping means alot when we do write expressions like this , thank you
Solution by the use of the squeeze theorem: We have lim x-> 0 of (1/sin(x) - 1/x) = lim x-> 0 of (x-sin(x))/(xsin(x)). We know that cos(x) < sin(x)/x < 1. Therefore (x-sin(x))/(xsin(x)) = (x-xsin(x)/x)/(x^2sin(x)/x) = (1 - sin(x)/x)/(xsin(x)/x) > (1 - 1)/(x*1) = 0. Similarly (x-sin(x))/(xsin(x)) = (1 - sin(x)/x)/(xsin(x)/x) < (1 - cos(x))/(xcos(x)) = (-1/cos(x)) * ((cos(x) - cos(0))/x). Now, the limits lim x-> 0 (-1/cos(x)) and lim x-> 0 (cos(x) - cos(0))/x both exist and they are finite (the latter is just the derivative of cos(x) evaluated at x = 0). We therefore have lim x-> 0 (-1/cos(x)) * ((cos(x) - cos(0))/x) = [lim x-> 0 (-1/cos(x))] * [lim x-> 0 (cos(x) - cos(0))/x] = -1 * (-sin(0)) = 0. It then follows from the squeeze theorem that lim x-> 0 (1/sin(x) - 1/x) = 0.
By the same reasoning: lim_x->0 (1/(x²+x) - 1/x) = lim (1/(x²+x) - (x²+x)/x(x²+x)) = lim(1/(x²+x) - lim(x²+x)/x lim(1/(x²+x)) = lim(1/(x²+x) - lim(1/(x²+x)) = 0 But the lim_x->0 (1/(x²+x) - 1/x) actually is -1. So your reasoning obviously is wrong somewhere.
Your reasoning is invalid because you may only use the product rule on limits if each limit is defined. The expression lim(1/sinx) does not have a well-defined limit so you are not allowed to split up the product like that
@@bjornfeuerbacher5514 Yeah, you're right, the reasoning is wrong. You cannot split the product of the limit like that unless both products are defined. In OP's example, lim(1/sin(x)) is not well-defined. In your example, lim(1/(x^2 + x)) is not well-defined, thus you cannot split the product of the limits like that
How is that enough? sin(x) - x goes to zero, right. But that does _not_ imply automatically that 1/sin(x) - 1/x also goes to zero. You have to do more work for showing that.
@@bjornfeuerbacher5514 the tild ~ means "equivalent" a(x) ~ b(x) means the limit of a(x)/b(x) tend to 1. And in practical term it means you can use one instead of the other without changing the limit, it's easy to see why.
What he shown is a rigorous method. Michael penn has done a video on a similar problem to show that the small angle approximation fails for the limit 1/x^2 - 1/sin^2 x where the answer is not 0. So no, you cannot just use the small angle approximation whenever.
since you have sinx - x divided by zero, the denominator may increase the small difference in the numerator, so you can’t ignore the fact that sinx isn’t exactly x. But you can do ( in this specific case) if you take in count the second grade approximation for sinx close to zero (see maclaurin): x-x^3/3! If you substitute this after the first expression transformation, the first grade x will simplifies, and you will have, as leading members, x^3 on numerator and a x^2 on denominator
@@robertveith6383 Well, it's standard to omit the "do this first" symbols when there is nothing even to be done first. Unless you just think everyone except you is writing math wrong, and fine, maybe you're right.
You made a mistake I believe with the second use of L'hospital's rule. You differentiated xcosx to x*-sinx + sinx * 1 when it should be x*-sinx + cosx * 1. So the final answer should be 0/2
Oh yea. Thanks for pointing it out to me.
Good thing it didn’t make a difference 😮💨
Second last step: (xcosx)'=-xsinx+cosx≠-xsinx+sinx
3:24 For the second usage of L'Hopital shouldn't the denominator be -x(sin x) + 2 cos x instead of -x(sin x) + sin x + cos x?
He made tge mistake tgere.
Yes, you are right. thank you for pointing that out!
@@bprpcalculusbasics blur lah you
Applying the engineering trick that sin(x) = x for small x, we can just do 1/x - 1/x = 0 and deal with it 😂
Chat, where I'm wrong?
Counterexample: x + x² = x for small x also holds. But nevertheless, 1/(x+x²) - 1/x does not go to zero. So no, this argument is not enough here.
@@Rozenkrantzz Err, my example _has_ the same condition. x + x² ~ x as x -> 0. You can easily calculate that for yourself: (x+ x²)/x = 1 + x, and for x -> 0, that goes to 1.
@@bjornfeuerbacher5514 Yeah I realize my mistake lol
@bjornfeuerbacher5514 they're kinda not the same, although this might be a little bit handwavy. x and sin(x) get arbitrarily close together as x approaches zero. But with x + x² that's kinda not the same, there's always this lingering small tail of x² that would make it different from 1/x at its limit. Or maybe not. But in any case, you're right, just saying that f(x) = g(x) at zero is not enough generally
@@arseniix For sin(x), there _also_ is such a "lingering small tail". If you look at the Taylor series of sin(x), that tail is x³/6.
The point is that for sin(x), there is no terms with x². _That_ is why the limit works for sin(x), but not for x+x².
And that's why your "proof" above is not enough, you have to invest more work.
For the sake of brevity, let lim f(x) denote the limit of f(x) as x→0.
lim(csc x - 1/x)
= lim ((x-sinx)/x³)(x/sinx)x.
Since the limit of each of these three terms exists, by algebraic limit theorem,
= [lim(x-sinx)/x³][lim(x/sinx)][lim x]
= ⅙(1)(0) = 0
This may be sloppy, but we know lim x-> 0 sin(x)/x = 1. So the two terms in the original are essentially equal near 0. So their difference is essentially 0 near 0.
Indeed 1/near 0 - 1/near 0. Can be anything.
Yes, that is indeed too sloppy. Counterexample: lim(x + x²)/x = 1, too. But nevertheless, 1/(x+x²) - 1/x does not go to zero.
After the first L’Hopital’s rule, we have
((1-cos(x))/x)/(cos(x)+sin(x)/x)
which has a limit if 0/(1+1) which is 0.
I see many comments mention the use of sin(x) ≈ x. Based on the video it intuitively cannot work because he used L'Hospital *twice* i.e. he differentiated twice, while sin(x) ~ x is a 1st order approximation (this is only an intuition, I never said it's a proof lol).
However if we consider the next term in the Taylor series, sin(x) ≈ x - x³/6. Thus we get
x/sin(x) ≈ x/(x - x³/6) = 1/(1- x²/6) ≈ 1 + x²/6
(In the last equality, use 1/(1-u) ≈ 1+ u)
In the end we get
1/sin(x) - 1/x ≈ 1/x + x/6 - 1/x = x/6 -> 0.
On the bottom we need cosx *1
Maclaurin is powerful when the variable is approaching 0
Let f and g be 2 functions such that f(x) ~g(x) when x->0. Does it means that 1/f - 1/g - >0 when x->0? No. Here it does but only by chance. Set f(x) =x and g(x) =x+x^p for some p>1. f~g as x->0, ok. 1/f-1/g = 1/x * (1-1/(1+x^{p-1})). But 1/(1+u) ~ 1-u when u->0 then 1/f-1/g ~ x^{p-2} (don't forget the 1/x) which goes to 0 if p>2, 1 if p=2, and even infty for 1
not addition or substraction.
You can do it without L'Hopital too ( Just for extra )
multiply divide by x+sinx in Nr and Dr
then make the equation as x^2 -sin^2x / xsinx ( x+Sinx)
rewrite it as x^2 - ( 1+cosx)(1-cosx)/xsinx(x+sinx)
Now take x^2 common from the Nr
and x from xsinx and another x from x+sinx
You'll get it done
Your post is wrong. You a missing needed grouping symbols in your numerator and your denominator.
@@robertveith6383 umm sorry sir i didn't get what you mean , it would be great if you tell exactly in what line , although i 've tried it on notebook that's how i told it up here , maybe the grouping of numerator and denominator on whole remains wrong here yeah maybe but you can definitely get what i mean ( although thanks alot ) Grouping means alot when we do write expressions like this , thank you
Honestly would've been kinda funny if 1 / sin(x) had been written as csc(x).
Solution by the use of the squeeze theorem:
We have lim x-> 0 of (1/sin(x) - 1/x) = lim x-> 0 of (x-sin(x))/(xsin(x)). We know that cos(x) < sin(x)/x < 1. Therefore (x-sin(x))/(xsin(x)) = (x-xsin(x)/x)/(x^2sin(x)/x) = (1 - sin(x)/x)/(xsin(x)/x) > (1 - 1)/(x*1) = 0. Similarly (x-sin(x))/(xsin(x)) = (1 - sin(x)/x)/(xsin(x)/x) < (1 - cos(x))/(xcos(x)) = (-1/cos(x)) * ((cos(x) - cos(0))/x). Now, the limits lim x-> 0 (-1/cos(x)) and lim x-> 0 (cos(x) - cos(0))/x both exist and they are finite (the latter is just the derivative of cos(x) evaluated at x = 0). We therefore have lim x-> 0 (-1/cos(x)) * ((cos(x) - cos(0))/x) = [lim x-> 0 (-1/cos(x))] * [lim x-> 0 (cos(x) - cos(0))/x] = -1 * (-sin(0)) = 0. It then follows from the squeeze theorem that lim x-> 0 (1/sin(x) - 1/x) = 0.
0:49 It's far faster to use Maclaurin.
We could use limit sinx/x = limit x/sinx = 1, as x goes to zero to solve this problem.
Could you please elaborate? How would be use that?
You made a BIG mistake with (xcosx)'
can we solve by using the approximation that : when lim x -->0 : sinx ~ x ??
as this would give us the answer instantly....
Counterexample: x + x² ~ x for x going to zero. But nevertheless, 1/(x+x²) - 1/x does not go to zero. So no, this argument is not enough here.
lim_x->0 (1/sinx - 1/x) =
lim (1/sinx - sinx/xsinx) =
lim(1/sinx) - lim(sinx/x)lim(1/sinx) =
lim(1/sinx) - lim(1/sinx) =
0
I think I did it right
By the same reasoning:
lim_x->0 (1/(x²+x) - 1/x) =
lim (1/(x²+x) - (x²+x)/x(x²+x)) =
lim(1/(x²+x) - lim(x²+x)/x lim(1/(x²+x)) =
lim(1/(x²+x) - lim(1/(x²+x)) = 0
But the lim_x->0 (1/(x²+x) - 1/x) actually is -1.
So your reasoning obviously is wrong somewhere.
Your reasoning is invalid because you may only use the product rule on limits if each limit is defined. The expression lim(1/sinx) does not have a well-defined limit so you are not allowed to split up the product like that
@@bjornfeuerbacher5514 Yeah, you're right, the reasoning is wrong. You cannot split the product of the limit like that unless both products are defined. In OP's example, lim(1/sin(x)) is not well-defined. In your example, lim(1/(x^2 + x)) is not well-defined, thus you cannot split the product of the limits like that
Use L.H rule direct in original funtion...and we get zero immediately🎉
The original version expression is not 0/0. It is infinity - infinity, which is not a condition for L’Hopital rule.
@@bobh6728 thanks
sin(x) ~ x at x -> 0. It’s enough.
How is that enough? sin(x) - x goes to zero, right. But that does _not_ imply automatically that 1/sin(x) - 1/x also goes to zero. You have to do more work for showing that.
@@bjornfeuerbacher5514js say TS expansion
@@davidbrisbane7206Could you please show some more steps? I know that limit, but I don't see immediately how one could use it to solve this problem.
@@bjornfeuerbacher5514
@1:48 you see
@@bjornfeuerbacher5514 the tild ~ means "equivalent" a(x) ~ b(x) means the limit of a(x)/b(x) tend to 1. And in practical term it means you can use one instead of the other without changing the limit, it's easy to see why.
inf-inf -> inf: ua-cam.com/video/ioEeRsT4-Sk/v-deo.html
曹老師你抄錯啦，第二次LH的分母那product rule抄錯了
啊 對吼 謝謝你
Could one use the fact that lim x->0 sin x / x = 1 for this? Intuitively it is clear that it causes a zero but how would one do it rigorously?
What he shown is a rigorous method.
Michael penn has done a video on a similar problem to show that the small angle approximation fails for the limit
1/x^2 - 1/sin^2 x
where the answer is not 0.
So no, you cannot just use the small angle approximation whenever.
since you have sinx - x divided by zero, the denominator may increase the small difference in the numerator, so you can’t ignore the fact that sinx isn’t exactly x.
But you can do ( in this specific case) if you take in count the second grade approximation for sinx close to zero (see maclaurin): x-x^3/3!
If you substitute this after the first expression transformation, the first grade x will simplifies, and you will have, as leading members, x^3 on numerator and a x^2 on denominator
You wrote that wrong. You are missing grouping symbols around the argument: sin(x)/x.
@@robertveith6383 Well, it's standard to omit the "do this first" symbols when there is nothing even to be done first. Unless you just think everyone except you is writing math wrong, and fine, maybe you're right.
haha cool
WRONG! WRONG! WRONG!
Usefully, sin(x) behaves like x at 0. How would we say this? Perhaps sin(x)∈Θ(x)
It's written sin(x) ~ x as x->0