Why Using L'Hopital's Rule is WRONG ⚠️

If you've used L'hopital's rule to solve this calculus limit problem, you've likely made the same mistake I did! Squeeze theorem is where L'hopital's rule comes from and so using it on this limit that is the famous squeeze thereom limit is a big calc 1 mistake...unless you understand why! Derivatives and indeterminate form make solving calculus AB limit problems like these a breeze with L'Hopital's rule, but it's inappropriate to use here. By the end of this video, you'll understand why. We'll use first principle on the formal definition of the derivative of sin x to demonstrate why L'hopital's rule is a trap!
0:00 - Introduction
0:26 - Plugging in X
0:32 - Indeterminate Form
1:00 - Using L'Hopital's Rule
2:34 - Sin x Derivative Proof (First Principle)

[@sunildhuri8421]

The L'H rule assumes that you know the derivative of the numerator and denominator but those derivatives are themselves obtained using first principle which uses limits .Wow!,I actually never thought of this.A good video by you highlighting that.

[@NumberNinjaDave]

Thank you

[@pedronunes3063]

Another option would be to define sin(x) and cossine(x) with the complex exponencial forms, right? So sin(x) = (e^xi - e^-xi)/(2i), cos(x) = (e^xi + e^-xi)/2

[@NumberNinjaDave]

But how does that solve the original problem of an indeterminate form?

[@pedronunes3063]

@@NumberNinjaDave If you use this definition (which is somewhat problematic, since it defines a real operation with complex numbers), the derivative of sine and cossine follow from the derivative of the exponencial.
In this case you can use L'Hôpital to calculate this limit, because it isn't circular reasoning anymore.

[@tokarak]

Using the claim that (L'Hopital's Rule is wrong) is not true. Neither is L'Hopital's Rule, in general, a redundant theorem, because it can be used for the division of functions which are antiderivatives of a known function. The title, description, and the premise of the video are all implying this.
As Memelord said, the sine-cosine functions can be defined through the taylor series; sin'(x)===cos(x) follows from quite easily from the "first principles" derivative of a polynomial. L'Hopital's Rule is fine, though it is not necessary if sin(x) is defined as a power series in x. In the other differential definitions of the trigonometric functions, L'hopital's rule becomes the most direct solution.
Breaking up the limit towards the end of a video makes the expression indeterminate, because the limits are infinite, and there is no single h variable any more.

[@NumberNinjaDave]

The Taylor series build off first principle indeed. But again, their prerequisite stems from already having used the first principle to begin with
The title isn’t to say that always using LH rule is wrong. It’s to say it’s wrong with the provided problem. The provided problem is the squeeze limit problem that was used to define LH Rule to begin with, so naively using LH rule in this case is simply wrong.
As for solving the problem provided, the Taylor series approach does show us one possible solution of 1 since when centering around 0, sin x converges to approximately x when x is super small
Of course, the other way to solve and prove is geometrically but not what this video is about.

[@Errenium]

​@@NumberNinjaDaveI have to disagree. If you define derivatives differently (e.g. using Grassman numbers), then L'Hôpital's rule does not depend on any limits

[@NumberNinjaDave]

@@Erreniumyou can disagree all you want. It’s circular reasoning to use LH rule on the same squeeze limit theorem problem that was used to define the rule in the first place. No need to define anything special to change the fact

[@asd-wd5bj]

@@NumberNinjaDave "It’s circular reasoning to use LH rule on the same squeeze limit theorem problem that was used to define the rule in the first place."
That's their point tho, you can just define it in such a way that doesn't depend on the squeeze limit theorem.
There are ways to define derivatives without limits (granted they are probably way past the level of your standard Calculus I course...), from which you can get the derivative of sin(x) without using the limit you're trying to solve

[@pastebee2812]

@@NumberNinjaDave It's not circular reasoning, where's the circle. It's just reasoning that doubles back on itself. "lim x->0 sin(x)/x = 1 by the squeeze theorem, therefore d/dx sin(x) = cos(x), therefore lim x->0 sin(x)/x = 1 by LH" is somewhat redundant (we already knew that limit from earlier) but entirely valid. It would be *circular* if you tried to prove it as "well d/dx sin(x) = cos(x), because lim x->0 sin(x)/x = 1 by LH, because d/dx sin(x) = cos(x), because lim x->0 sin(x)/x = 1 by LH, because [...]", because the argument keeps going forever, and proofs have to be finite. If going "downwards" through the proof terminates at the squeeze theorem, then it's finite and therefore valid.

[@weekendfriday22]

really clear , good video

[@NumberNinjaDave]

Thanks so much

[@danielc.martin1574]

Define trig functions via the integral of area of circle or perimeter of circle and its okay to use the hospital

[@NumberNinjaDave]

Yup, if people understand that geometric proof methodology for proving squeeze theorem first and actually understand where LH rule comes from, then yes

[@ntlake]

It depends on how you define sin(x) and cos(x). If you define them using Taylor series, then you can easily prove that the derivative of sin(x) with respect to x is cos(x) and so you can use l'Hopital's rule. At that point, however, it's much simpler to write sin(x)/x = (x + o(x))/x, which clearly approaches 1 as x approaches 0.
AIso, in order to be able to apply l'Hopital's rule, a necessary condition is that the limit of f'(x)/g'(x) exists, which is not true in general. Take f(x) = x + sin(x) and g(x) = x as a counterexample: the limit f(x)/g(x) as x approaches infinity is an indeterminate form which can be easily shown to approach 1. However, if you try and apply l'Hopital's rule, you get the limit as x approaches infinity of f'(x)/g'(x) = 1 + cos(x), which doesn't exist.

[@NumberNinjaDave]

All valid points! However, keep in mind that Taylor series stem from the same condition of knowing the nth derivative and the main takeaway of this video as mentioned in the last few seconds was to understand the derivation and usage of LH rule before just blindly using it. Students should know how to solve this squeeze theorem limit without LH rule. Otherwise, it’s just blind formula memorization and I teach understanding, not memorization.

[@ntlake]

@@NumberNinjaDave I didn't say to use Taylor series, I said to *define* sin(x) with its "Taylor series". That is, define sin(x) = x - x²/2 + x³/6 - x⁴/24 + ...
This means that you don't need derivatives anymore to get to that power series, since you made it true by definition. Many papers prefer to adopt the power series definition or the integral definition for some functions, since it often simplifies calculations.

[@NumberNinjaDave]

@@ntlake and again, that definition comes from derivatives. You are solving for the constants needed to satisfy an infinite term Taylor series. Go look up where the specific Taylor series definition of the sin function comes from.

[@ntlake]

@@NumberNinjaDave it's not a Taylor series anymore, it's a power series definition, even though it's often called "Taylor series definition" because that's where the idea came from. It makes no sense at all to say "that definition comes from derivatives", a definition doesn't come from anything. It's a definition.
Defining sin(x), cos(x) and many other functions through power series is common practice in modern calculus. One can then show that the power series definition is consistent with their Taylor series, but that's another thing.

[@Alexj_movieguy]

so you would have to use squueze theorem to solve the sinh/h limit? so in that case just solve the original limit using squeeze theorem? that's what i gathered.

[@NumberNinjaDave]

Bingo!

[@cucler6718]

I'm not sure if I do it correctly
To derivate sin(x)
Lim h>-0 (Sin(x+h)-sinx)/h
Lim h->0 (sinxcosh+sinhcosx - sinx)/h
Now I consider the sine of h to be h since h Is almost 0 and sine of 0 is 0
And cosine would become 1.
Lim h->0 (sinx + hcosx - sinx)/h
Lim h ->0 hcosx/h
So cancelling the h you get cosx

[@NumberNinjaDave]

Taylor series shows us just that and squeeze theorem, that when centered around 0 and for very small x, sin x is roughly x. But I don’t use Taylor series here since that too uses the definition of the derivative already

[@cucler6718]

@@NumberNinjaDave so... Is it the correct way or not?

[@NumberNinjaDave]

@@cucler6718 your math looks correct and gets the right answer…question for you though. Do you think that approach is the right approach?

[@cucler6718]

@@NumberNinjaDave I really don't think it's the right approach lol

[@NumberNinjaDave]

@@cucler6718 agreed

[@stevepreston7879]

Who said you can split a limit of sums into a sum of limits? In this case you ignored the first and last limits, which are undefined. So I claim that step is unjustified.

[@NumberNinjaDave]

You can definitely split them. Yeah, I definitely ignored them since the middle one was sufficient on its own to prove the point of the video

[@NumberNinjaDave]

@@berryesseen infinity isn’t a number so we aren’t changing the limit. It’s unbounded. You certainly can.

[@NumberNinjaDave]

@@berryesseen if you say so 😉 you can disagree all you want. Looking at your channel, it looks like you’re bashing my channel in an attempt to draw viewers to your channel and to troll. That tells me everything I need to know.

[@NumberNinjaDave]

Quiz: LH rule is an awesome rule to simplify limits, but for the squeeze limit theorem limit from this video, why is it incorrect to use it without caution?