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.
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.
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
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.
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.
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).
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.
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.
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).
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.
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!!
![lim [x-tanx]/[x-sinx] as x approaches 0](/img/logo.svg){kind=logo}
