Indeterminate form (or 0/0) -> do more work to see if you can factor the denominator out of the numerator. Works most of the time - a current Calc student who struggled with this stuff a lot

jus do the taylor series and realise youre not gonna get any constant terms but you are gonna get an x term so its gonna go to infinity

Surely L'Hopital's rule is the fastest way here, since the numerator and denominator are both so easy to differentiate.

Ah, this takes me back to my A Levels. Thanks for the trip down memory lane, I wish I had your videos when I was a student!

Let f(x) = x and its domain D of f(x) be { x in R } Let g(x) = x^2/x and its domain D of g(x) be { x =/=0 } or { x in R\{0} } What you are highlighting here is that for two functions to be considered truly equal, their domains must be preserved and the same. If at some point the domain has changed via simplification then the two are no longer equivalent expressions. Now with this being said, I have the following questions (with my own commentary) if you don't mind answering: 1. Genuine question and this is not me trying to be obtuse, but what is the practical benefit of preserving a hole (I refer to the 'hole' as being punched in at the origin) in such an expression? As far as I can see there is very little benefit to thinking of expressions in this manner. I say this purely from a practical perspective wherein the application of Maths to be carried out is say for some Engineering/Physics/Economics style problem, then this is of no use conceptually (at least on the surface). 2. Why does it matter if one actually changes the domain via simplification? I say this because if one considers the example of { g(x) = x^2/x | x in R\{0} }, it feels like this 'problem' has been very particularly manufactured to be an issue. Additionally, I could arbitrarily restrict the domain for f(x) like doing: {f(x) = x| x in R\{0} } and then the domains for f(x) and g(x) would be equal and therefore actually be the same. Now clearly me arbitrarily removing zero from the domain of f(x) is silly, but it gives a virtually similar effect of not simplifying the expression for g(x) to its simplest form. I could also just pick intervals that never involve zero ever so that the two expressions are the same.

I was thinking on converting the function into a taylor series

I did it basically the same way. Except I did a substitution, avoiding the need for the chain rule: L = lim<x~+inf>{ xx sin(1/x) } let y = 1/x L = lim<y~0+>{ sin(y) / yy } sin(0) = 0 0^2 = 0 (d/dy) sin(y) = cos(y) (d/dy) yy = 2y L = lim<y~0+>{ cos(y) / 2y } = +inf

Or you can just devide and get the 1-1/(x+2)

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Be careful with those parentheses :)

What i would do is to make a variable change θ=1/x, x->0 => θ->∞ so we have the limit when θ->∞ of sinθ. Now we do know that -1≤sinθ≤1 so we take the limit in all 3 parts of the inequality, and the limit of -1 and 1 is -1 and 1, respectively so the limit oscilates between -1 and 1 so the limit diverges

i don't really get why americans use l'hospital's rules so often when using basic asymptotic analysis works much better and isn't much of a leap in terms of theory

I was thinking integral by parts

Lim x->♾️ (xsin(n/x) = n (Taylor Series) so the answer gets oblivious from there.

x^2 sin(1/x) = x (sin(1/x)/1/x) let x = 1/t x tends to inf then t tends to 0 therefore, (1/t)*(sin(t)/t) =1/t (as sinx/x =0 if x tends to 0) = inf (not partial limit sub btw if i am correct)

An elementary proof that doesn’t require fancy theorems : Our starting point, for all numbers within ]0,1[ : sin(x)>x-x^2 (This classic inequality can be proven through differentiation) Thus, by substituting x with 1/x, for all numbers greater than 1 : sin(1/x)>1/x-1/x^2 We finally obtain : x^2*sin(x)>x-1 As x approaches ♾️, x-1 does too, so by the lower bound theorem, x^2*sin(x) also approaches infinity.

Another method I'd like to use when solving this problem let 1/x = k, so k approaches 0. So we can rewrite the equation as Limit:k-->0 (1/k²)(sin k) Limit:k-->0 (sin k/k²) using the squeeze theorem we can simplify the equation to Limit:k-->0 (1/k) and finally 1/0 which is infinity

isn't the limit of a product the product of the limits iff both limits exist?

as x approaches infinity, 1/x becomes an exponentially smaller number. small angle approximation for sin(u) where u is small, sin(u) ~= u. what remains is x^2 * 1/x = x. lim_x-->infinity (x) = infinity

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x=sin(teta) , y=cos(teta) , or , x=cos(teta) , y=sin(teta) ,

neat.

Explaining discontinuity with this creativity is amazing, your content is great!! Keep it up 🗣️🔥

Hmm, can't you let d/dx abs(x) be sign(x)? Then product rule gives you |x| + x sign(x) (everywhere except at zero!) but then x sign(x) is |x| again, so you get 2|x| everywhere except zero, and restore the value at zero by continuity if you insist the derivative must be continuous. (it's certainly true that there's a unique continuous function that's the derivative of x |x|, but there are lots of almost everywhere continuous functions that are almost everywhere the derivative of x |x|)

cant you just also define |x| as the squareroot of x^2, in which case the derivative would be (2x^2)/|x|? in this case, x=0 is undefined, but the limit as x approaches zero is 0.

Sin(x) does not equal 1 at 3pi/2 it is equal to -1.

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man calculus is easy enough, you should be teaching US how to get a wife 😂

One of the most remarkable parts of the power rule integral to me, is how the special case of n = -1 where it fails, can ultimately be shown to still be consistent with the power rule. Simply let h be an infinitesimal offset from -1, such that n = -1 + h. Then strategically assign the constant of integration to -1/h. integral x^(-1 + h) dx = 1/h * x^h - 1/h Regroup: (x^h - 1)/h Then take the limit as h goes to zero. Rewrite the power term using natural log: x^h = e^(ln(x)*h) Thus: (e^(ln(x)*h) - 1)/h Use L'H's rule, to take care of the 0/0: dN/dh = ln(x)*e^(ln(x)*h) dD/dh = 1 Thus the limit is: ln(x)*e^(ln(x) * h) Plug in h = 0, and we see the exponential term becomes 1. The result becomes ln(x), exactly as we expect.

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I hope this made implicit differentiation easier! If you enjoyed this video, check out some goodies and helpful tools below! 🛍 Want some 🥷🏿 swag? Shop my goodies HERE! tinyurl.com/numberninjaswag *********************************************************** 📚Helpful stuff and my favorite MUST haves I used in my college courses ⬇ Math and school making you anxious? I totally get it and wrote this book for YOU: amzn.to/3Y2LWKv Here’s a great study guide so you can CRUSH your AP exam, like a ninja! amzn.to/3N5pjPm This graphing calculator is a beast and never failed me in college: amzn.to/4eBNeRS I loved THIS ruler in college, for engineering classes: amzn.to/4doupRk These are my affiliate links. As an Amazon Associate I earn from qualifying purchases. ***************************************************************************************

L'Hopital's rule is circular series expansion doesn't help because derivative is still needed

I spend so much time goofing up the Algebra that I scarcely stop to think about the meaning behind these equations. Calculus is the easy part, imo, and shows more holes in my knowledge (e.g. rational expression manipulation) than it does for the graphs. Excellent video, though! Straight and to the point, easy to follow, with a bite-sized couple kernels of thought-provoking additives.

I think that when she asked what d is she hadn’t yet understood that d itself isn’t a number, it’s a symbol,that stands for a change that’s happening. A symbol saying there’s a difference in a number, but d is not a number itself.

great video man i like it and i like mario

Well, your dimensions don’t work out. 2 cows/month x 1 month = 2 cows. Month cancels.

Unfortunately intuition takes a while to get. Some people use repetition to etch intuition for a topic into their minds, this has too many flaws to list here. It is a "Shortcut" for intuition. A way to get to the end of the tunnel, when you dont know which way you are facing. The other is to understand it perfectly such that you have no doubts. This is a much more powerful method compared to repetition but it requires a lot more active thought and time put into the topic. It's powerful because you not only understand a topic easily, you also develop a wide range of different but interconnected skills with the topic in hand. What this video attempts to do is a mix of both while leaning heavily towards the latter, which is neither necessarily wrong nor is it correct but the second method of learning requires quite a lot of energy put in, which just isn't possible for a normal person within the span of a few minutes. They'd be better off learning from the absolute basics, rather than put 5 minutes here. Do not get me wrong. This is in no way a waste of time but the timeframe utilised is just too small to make a enough difference to justify putting this video over a more in depth one. The other issue is that calculus itself is a really massive branch of mathematics. Limits, Derivatives, Integration and continuity are all very important parts of calculus that are interconnected to each other. It is not possible to understand derivatives and continuity without derivatives and it is not possible to understand integration without derivatives.

I like your use of the term "stitch together". However, I think you spent too much time on non-essentials (all the business about units, area of triangle, etc) and yet tossed in terms like "limits of integration" with barely any explanation at all. Even a brief mention of where the integration symbol comes from (stylized "S" for "sum") would help. Likewise, the "dx" part. Your wife kept asking about "d". Again, you introduced a term without really explaining it (in simplest terms, anyway) - more of a hand-wave - and still didn't really answer her question. Wouldn't it have been simpler to go back to Riemann sum (aka rectangle rule) form, showing f(x) times delta(x)? For someone familiar with simple geometry, it should be fairly painless to go from making delta(x) smaller and smaller to get better approximation, to "now imagine if delta(x) was infinitesimally small - a single point", and there are "infinitely many" of them that we're now going to "stitch" together" which will give us not just a very good approximation, but an *exact* answer. THAT's calculus! :)

Yeah but still i don't think my wife would understand 😥

Nice video!

What I learned from this video: 5 minutes is not enough time to understand calculus.

I think you did not show how the area is the number of cows. Maybe trying it out by examples would help convince a beginner.

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Let x=2Sinz dx = 2coszdz Int 2coszdz/ 2cosz Int dz z+c arcsin(x/2)+ c

Piecewise continous

The first method is not wrong precisely. since there is no function f(x) such that f(x)=y for all x,y on the circle, we don't really have anything to define the derivative as. in a formal maths what you would do is create isolated functions (by the implicit function theorem, they exist) that match the original equation and the implicit function theorem gives you the derivative (which, you would not believe this, comes to -x/y). so, neither of these methods are formally correct. they both give you a right answer (if you manage the +- carefully you will get -x/y). that's not to say the 2nd method is bad. it's actually good, because it dosen't require any square roots, which are easy to mess up. excellent video either way, shows a good solution to the problem without wasting your time.

Hey, nice video and clear explanation! I would assume y cannot be zero because it is the denominator of the fraction. That makes sense, the tangent at y=0 would be vertical.