It's so cool now that I'm at university that I get to understand every little detail of each theorem and algebraic manipulations we can do in order to solve those tricky limits. It's so fun that I get to see a problem, try it by myself and then be shown a slightly different approach which really boosts learning things, i've been a digital student of yours since a long time, thank you for all your older and newer content :D
5:05 Pausing at this point to say that my instinct after two applications of l'hoptial's rule, is just to cancel out the sin(x) inherent in tan(x), leaving -2sec^3(x), which is -2, by direct substitution.
Taylor series expansion: x-tanx=x-(x+x^3/3+2x^5/15+...)=-(x^3/3+2x^5/15+...) Taylor series expansion: x-sinx=x-(x-x^3/6+x^5/120-...)=x^3/6-x^5/120+...) Divide the numerator and denominator by x^3: (x-tanx)/(x-sinx)=-(1/3+2x^2/15+...)/(1/6-x^2/120+...) The limit as x→0 is -(1/3)/(1/6)=-2
This only works if you are taking the limit around x=0. It's actually equivalent to l'hôpital's rule because it is effectively taking repeated derivatives. But it's the way I would have approached it.
Actually at the step you quit using La Hospitals rule you could have simply substituted sinx/cosx for tan x and simply cancel the problematic sin x and reach the same answer.
Immediate with a limited developpement. tan(x) = x + (1/3).(x^3) +o(x^3) and sin(x) = x + (-1/6).(x^3) + o(x^3) Then (x - tan(x)) / (x - sin(x)) = (1/3)/(-1/6) + o(1), then the limit is -2
5:32 You can apply L'Hôpital's rule as many times as you like. Because you only have sec(𝑥) and tan(x) in the numerator, the derivatives will also contain only those functions. And the denominator will just switch between sin(𝑥) and cos(𝑥). So, at some point you'll have rewrite the numerator in terms of sin(𝑥) and cos(𝑥).
As already notice, I think that Taylor expansion is much quiches for the same result. But it is à very good exemple to demonstrate that Hospital rôle is not a panace for the limit 0/0.
After one application of L'Hopital's Rule, I noticed that the numerator simplified to -tan^2(x). So I multiplied the numerator and denominator by 1+cos(x) to make the denominator sin^2(x). That left me -(1+cos(x))/cos^2(x), which evaluated simply to -2. A bit more complicated trigonometrically, but a bit simpler in the calculus, I think.
This also gives us an interesting way to approximate pi! (Space to hide the final result of the video) If we say (x-tan(x))/(x-sin(x))=-2, which it approaches as x->0, then we can re-arrange 2*sin(x)-2*x=x-tan(x) -> 3*x=2*sin(x)+tan(x) x approaches 2/3*sin(x)+1/3*tan(x). And because for simple angle fractions of pi, like one fourth and one twelfth, are easily constructable with the right polygon, the length of sin(x) and tan(x) will be too. So n*[2/3*sin(pi/n)+1/3*tan(pi/n)] -> pi How good is this approximation? Well, for n=6, we are off by one part in 240. For n=12, we are off by one part in 4,153. That's not bad! There are other, faster approximations of pi, but few can be done with such ease.
Im not sure if it is the early morning or the last limit video he did, but I didnt even think about using L'Hopitals rule initially. But after watching, L'Hopitals rule twice then simplifying doesnt seem so bad
As Newton kind of explains in the video, this problem actually screams L'Hôpital! The second derivative of 𝑥 is zero and the second derivative of a trig function of 𝑥 (not including the inverse trig functions) is at worst going to be a product of trig functions of 𝑥, possibly with an integer factor as well. So, applying L'Hôpital's rule twice and simplifying, we are bound to end up with a rational expression in sin 𝑥 and cos 𝑥, with any sin 𝑥 factors being in either the numerator or the denominator. If the sin 𝑥 factors are in the numerator, then the limit as 𝑥→0 is zero. If the sin 𝑥 factors are in the denominator, then the limit is either +∞ or −∞. If there are no sin 𝑥 factors, then the limit is going to be a rational number.
I just used l’Hopital’s rule I think 4 times. The difference is that I ignored sec and just took the derivative of tan(x) = 1+(tan(x))^2.This way the chain rule derivations aren’t that bad.
I did it in much the same way, except that as soon as I had 1 - sec^2 x, I converted it into a fraction and factored 1 - cos^2 x into (1 - cos x)(1 + cos x).
sir i don't understand at the part where you were cleaning up -2secx(secxtanx)/sinx that why we ended up with -sec^2xtanx/sinx instead of -2sec^2xtanx/sinx
I spend over 3 years trying to solve this limit without using lohpital's theorem, but I faild every time.... I will be greatful if you solved it another way not include series❤
You'd just get the answer in 3Rd step bro Secx . Tanx =sinx It would simply cancel out denominator And u would have -2secx with you and limx----0 -2secx equals simply -2
Whenever you say another person's work is incorrect, always provide the flaw in the work or provide tour solution. I will come back to this comment after a while to see your response.
At 1:55 terrible handwriting in excellent handwriting on chalkboard error!? 😮 If only I could write on chalkboards so clear. ... I agree! I knew the L'Hopital's Rule had to be used once as (x - xsinx/cosx)/(x - sinx) became too cumbersome to instantly factor. You could have, noticing in L'Hopital's rule you had done twice, that the tanx that is sinx/cosx around x=0 has sinx in the numerator multiplied by your sec^x/denominator sinx and multiplied to another secx. ... Everything easily became -2(secx)^2 times another secx leftover!? From your second taking derivatives by L'Hopital's Rule just as easy as a step as factoring 1 - secx after only the 1st derivative and then making another step common denominator and cancelling numerator and denominator same terms. It's called "poles and zeros cancellations" in Electrical Engineering because whether it is sinx/sinx or (1 - cosx)/(1 - cosx) we know we are cancelling a numerator zero "called zero" with a denominator zero "called pole" like it were a multiple by cancellation "one" times the rest of the non cancelled products in numerator and denominator. By the way without putting the limit of x=0 evaluation in on your derived avoidance to a second derivative showed L'Hopital's Rule 2nd derivative result ... I think!? I have to figure what went wrong ... Sort of proof: -(1 + secx)/cosx = -secx(1 + secx) = -(secx + (secx)^2) = -(cosx + 1)((secx^2)). !? That somehow is -2(secx)^3!? Math went wrong!?! 😭 😂
@@canyoupoop ... The limit is the same, I agree! But -2(secx)^3 I don't think is -secx(1 + secx) in leftovers on the left to right sides!!? They both evaluate as -2 with x = 0. I am rechecking what went wrong in the math for when x not 0. 🤯
@@canyoupoop ... Further thoughts! The results of a new function g"(x)/f"(x) is no longer the function g'(x)/f'(x) ... L'Hopital's Rule is only good at x = 0 and cannot be applied to any numerical numerator/numerical denominator not of indeterminate forms. 2secx^3 is not supposed to equal secx(1 + secx) because the g'(x)/f'(x) new function was not meant to be g(x)/f(x) where L'Hopital's Rule is the composite function of tangents ratios and not at all being the representing function of the function anymore. That is how one side divided out sinx/sinx and the other side divided out (1 - secx)/(1 - sinx). ... So it is what it is at x=0 only! Not for any arbitrary x relationship factoring out different factors multiplying with the formed 0/0 identical numerator and denominator same forms of 0/0 or +/-infinity/infinity!!
We also can write tanx/sinx =1/cosx and then it will become -2sec^3x and answer will be -2
I was screaming that at my phone that time
@@canyoupoop Me too
I was thinking that. But after applying L'Hopital rule, replacing secx with 1/cosx was easier as pure algebra follows from there
How'd you learn all these 😭😭
It's so cool now that I'm at university that I get to understand every little detail of each theorem and algebraic manipulations we can do in order to solve those tricky limits. It's so fun that I get to see a problem, try it by myself and then be shown a slightly different approach which really boosts learning things, i've been a digital student of yours since a long time, thank you for all your older and newer content :D
5:05 Pausing at this point to say that my instinct after two applications of l'hoptial's rule, is just to cancel out the sin(x) inherent in tan(x), leaving -2sec^3(x), which is -2, by direct substitution.
Exactly my point
Exactamente buena observación
Taylor series expansion: x-tanx=x-(x+x^3/3+2x^5/15+...)=-(x^3/3+2x^5/15+...)
Taylor series expansion: x-sinx=x-(x-x^3/6+x^5/120-...)=x^3/6-x^5/120+...)
Divide the numerator and denominator by x^3: (x-tanx)/(x-sinx)=-(1/3+2x^2/15+...)/(1/6-x^2/120+...)
The limit as x→0 is -(1/3)/(1/6)=-2
This only works if you are taking the limit around x=0. It's actually equivalent to l'hôpital's rule because it is effectively taking repeated derivatives. But it's the way I would have approached it.
@@FadkinsDiet yeah by LH rule i just solved it in 3 lines
Actually at the step you quit using La Hospitals rule you could have simply substituted sinx/cosx for tan x and simply cancel the problematic sin x and reach the same answer.
Immediate with a limited developpement. tan(x) = x + (1/3).(x^3) +o(x^3) and sin(x) = x + (-1/6).(x^3) + o(x^3)
Then (x - tan(x)) / (x - sin(x)) = (1/3)/(-1/6) + o(1), then the limit is -2
5:32 You can apply L'Hôpital's rule as many times as you like.
Because you only have sec(𝑥) and tan(x) in the numerator, the derivatives will also contain only those functions.
And the denominator will just switch between sin(𝑥) and cos(𝑥).
So, at some point you'll have rewrite the numerator in terms of sin(𝑥) and cos(𝑥).
As already notice, I think that Taylor expansion is much quiches for the same result. But it is à very good exemple to demonstrate that Hospital rôle is not a panace for the limit 0/0.
After one application of L'Hopital's Rule, I noticed that the numerator simplified to -tan^2(x). So I multiplied the numerator and denominator by 1+cos(x) to make the denominator sin^2(x). That left me -(1+cos(x))/cos^2(x), which evaluated simply to -2. A bit more complicated trigonometrically, but a bit simpler in the calculus, I think.
This also gives us an interesting way to approximate pi! (Space to hide the final result of the video)
If we say (x-tan(x))/(x-sin(x))=-2, which it approaches as x->0, then we can re-arrange
2*sin(x)-2*x=x-tan(x) -> 3*x=2*sin(x)+tan(x)
x approaches 2/3*sin(x)+1/3*tan(x). And because for simple angle fractions of pi, like one fourth and one twelfth, are easily constructable with the right polygon, the length of sin(x) and tan(x) will be too. So n*[2/3*sin(pi/n)+1/3*tan(pi/n)] -> pi
How good is this approximation? Well, for n=6, we are off by one part in 240. For n=12, we are off by one part in 4,153. That's not bad! There are other, faster approximations of pi, but few can be done with such ease.
thank you for this man love you.
I asked for trigonometry before and you gave.
Fantastic! NEVER STOP LEARNING!!!!!
Im not sure if it is the early morning or the last limit video he did, but I didnt even think about using L'Hopitals rule initially. But after watching, L'Hopitals rule twice then simplifying doesnt seem so bad
As Newton kind of explains in the video, this problem actually screams L'Hôpital!
The second derivative of 𝑥 is zero and the second derivative of a trig function of 𝑥 (not including the inverse trig functions) is at worst going to be a product of trig functions of 𝑥, possibly with an integer factor as well.
So, applying L'Hôpital's rule twice and simplifying, we are bound to end up with a rational expression in sin 𝑥 and cos 𝑥, with any sin 𝑥 factors being in either the numerator or the denominator.
If the sin 𝑥 factors are in the numerator, then the limit as 𝑥→0 is zero.
If the sin 𝑥 factors are in the denominator, then the limit is either +∞ or −∞.
If there are no sin 𝑥 factors, then the limit is going to be a rational number.
I just used l’Hopital’s rule I think 4 times. The difference is that I ignored sec and just took the derivative of tan(x) = 1+(tan(x))^2.This way the chain rule derivations aren’t that bad.
I did it in much the same way, except that as soon as I had 1 - sec^2 x, I converted it into a fraction and factored 1 - cos^2 x into (1 - cos x)(1 + cos x).
Nice class sir❤
Very good. Thanks 🙏
what exactly is sec(x) ? I never learned that at school --' I only learned sin(x), cos(x), tan(x) and cot(x)
sec(x) = 1/cosx
cosec(x)=1/sinx
@@SankalpaSatyal thanks mate !
Awesome video sir, can we solve it without using bernoulli's rule??
Why you saying Bernoulli rule, he literally sold it. It's not Bernoulli's anymore it belongs to L'Hôspital
sir i don't understand at the part where you were cleaning up -2secx(secxtanx)/sinx that why we ended up with -sec^2xtanx/sinx instead of -2sec^2xtanx/sinx
im asking this because my approach here was L'h
I would have just divided everything by x. We know the limit of sin(x)/x equals 1. And tan(x)/x=sin(x)/x*1/cos(x).
Genial
Simplifie by sinx in the 2rd hopital rule ...
Using the taylor series this gets much easier.
What about -2 onsecx(secxtanx)
I spend over 3 years trying to solve this limit without using lohpital's theorem, but I faild every time....
I will be greatful if you solved it another way not include series❤
I think some limits are beyond just algebra. Calculus must be employed.
Sir why is it not -2secx alone why -2secx(secxtanx)?
You'd just get the answer in 3Rd step bro
Secx . Tanx =sinx
It would simply cancel out denominator
And u would have -2secx with you and limx----0 -2secx equals simply -2
5:19 you missed number 2
First, the answer is incorrect. Second, don't you know the Equivalent Infinitesimal? 1-sec²x=-tan²x~-x² , 1-cosx~x²/2。
Whenever you say another person's work is incorrect, always provide the flaw in the work or provide tour solution. I will come back to this comment after a while to see your response.
At 1:55 terrible handwriting in excellent handwriting on chalkboard error!? 😮 If only I could write on chalkboards so clear. ... I agree! I knew the L'Hopital's Rule had to be used once as (x - xsinx/cosx)/(x - sinx) became too cumbersome to instantly factor.
You could have, noticing in L'Hopital's rule you had done twice, that the tanx that is sinx/cosx around x=0 has sinx in the numerator multiplied by your sec^x/denominator sinx and multiplied to another secx. ... Everything easily became -2(secx)^2 times another secx leftover!? From your second taking derivatives by L'Hopital's Rule just as easy as a step as factoring 1 - secx after only the 1st derivative and then making another step common denominator and cancelling numerator and denominator same terms. It's called "poles and zeros cancellations" in Electrical Engineering because whether it is sinx/sinx or (1 - cosx)/(1 - cosx) we know we are cancelling a numerator zero "called zero" with a denominator zero "called pole" like it were a multiple by cancellation "one" times the rest of the non cancelled products in numerator and denominator. By the way without putting the limit of x=0 evaluation in on your derived avoidance to a second derivative showed L'Hopital's Rule 2nd derivative result ... I think!? I have to figure what went wrong ... Sort of proof: -(1 + secx)/cosx = -secx(1 + secx) = -(secx + (secx)^2) = -(cosx + 1)((secx^2)). !? That somehow is -2(secx)^3!? Math went wrong!?! 😭 😂
What you mean, plug x=0 is sec³(x) and your limit is indeed -2
What do you mean?
@@canyoupoop ... The limit is the same, I agree! But -2(secx)^3 I don't think is -secx(1 + secx) in leftovers on the left to right sides!!? They both evaluate as -2 with x = 0. I am rechecking what went wrong in the math for when x not 0. 🤯
@@canyoupoop ... Further thoughts! The results of a new function g"(x)/f"(x) is no longer the function g'(x)/f'(x) ... L'Hopital's Rule is only good at x = 0 and cannot be applied to any numerical numerator/numerical denominator not of indeterminate forms. 2secx^3 is not supposed to equal secx(1 + secx) because the g'(x)/f'(x) new function was not meant to be g(x)/f(x) where L'Hopital's Rule is the composite function of tangents ratios and not at all being the representing function of the function anymore. That is how one side divided out sinx/sinx and the other side divided out (1 - secx)/(1 - sinx). ... So it is what it is at x=0 only! Not for any arbitrary x relationship factoring out different factors multiplying with the formed 0/0 identical numerator and denominator same forms of 0/0 or +/-infinity/infinity!!
@@lawrencejelsma8118 Oh nice
😅😅 thanks
🇵🇸