I keep explaining Conjugates all the time and how important they are, this is the first time I've seen use of the conjugate to "unrationalize" the denominator, but to great benefit.
Yeah i mean, before l'hopital/hospital was introduced this was the method we were taught. I did mentally using l'hopital. Idk if more steps = more value to getting to the same answer 😮
Correct me if I'm wrong but wouldn't plugging in any number that approaches 1 immediately get you the answer as well? Numbers that approach 1 from both sides like 0.999 or 1.001 should get you either -2.0001 or -1.9999.
(1-sqrt((2x^2)-1))(1+sqrt((2x^2)-1)/(x-1)(1+sqrt((2x^2)-1) equal to 1-2x^2+1/number (too lazy to write) 2-2x^2/numb which factor that to get 2(1-x^2) or -2(1+x)(x-1)/(x-1)(1+sqrt((2x^2)-1) cancel out -2(x+1)/(1+sqrt((2x^2)-1) lets try it -4/2 or -2 boom the answer is -2
Hey Andy. This mai not be that exciting (or maybe it will). I saw this on the news and they say that this is a 4th grade problem in my country (i highly doubt it): "what is the lowest natural number that if you take the last digit from that number and move it to the front of the number, the resulting number would be the double of the original number" (Something like: 2 x abc = cab. Or 2 x abcd = dabc. You get the picture) The result is going to surprise you 😁
@@mathprofessorpk1234 bro is there any limitations to this method Because this looks so easy and efficient Then why didn't no one had ever told me about this?
I like L'Hopital's rule but another way to solve is the following: this looks like one root of the quadratic formula for ay^2 + by + c = 0 with a = (x-1)/2, b = -1, and c = -x-1. As x ->1, a -> 0 and this becomes a linear equation -y-2 = 0 => y = -2
Well, since we cant apply a direct substitution we are going to need to manipulate the expression. Since there is a radical in the numerator i suppose we are going to multiply by the conjugate, which is a fancy way of multiplying with 1. The conjugate is the numerator with opposite signs. So change the subtraction to addition. Multiply the top and bottom. Now we need to look at what we have and see if we can do anything like finding the difference of squares or cubes, we need to see if we can factor so we can remove as much terms as possible so we can do a direct substitution. Just follow your axioms and theorems. Its not hard with practice.
I would’ve stopped at 0/0 undefined, how does one know when to go further? What about the other times when we end up we undefined, is there an answer with a different approach? Uhmm, this seems like math vodoo now.😮
I love limits and calculus but since learning l’hopitals rule at home, I’m never not using it (except when my calc teacher says that I can’t use it because our class hasn’t covered it yet)
Wait, i don't understand by just giving more work you got the correct answer? So just putting the values were not the correct way because it gave you 0/0?
Usually 0/0 is an error because we can’t divide by 0 so we find the limit: using values closer and closer to approach the number we wanna input and check what it gets closer to. But mathematicians realized instead of inputing so many closer values we can just use algebra and cancel the terms to get it’s limit. Ik yeah this is crazy how could something be wrong but algebraically correct?? It makes sense once you see the proofs
So a limit is asking what the value of the function is *approaching* as the x values approach (in this case) 1, not what the value of the function *is* when the x value *is* 1. If the function is continuous, then the approaching value and the ‘is’ value are the same thing, so using direct substitution works. But this square root function is not continuous because of the denominator. When we ‘do more work’, what we are doing is manipulating one function into a slightly different function: actually it is the same function at all points of the original except that it is now continuous at the limit value, so we can use direct substitution on the simplified function. The y values around x = 1 will be the same for both functions so the limits will be the same.
3(x-5)²=-10 My Teacher Gave us this and until now everyone including the teacher herself can't solve it. Woulf you like to try and solve this through Quadratic Formula? -b±√b²-4ac/2a
Dude you could've done it way simpler if you took: f(x)=-sqrt(2x^2-1) Then the limit would be f'(1) So all you need to do is evaluate the derivative of f(x) at x=1 and there is the result
And if you want to be cheeky and give your math teacher a headache, read forward to the next chapter of your textbook to realize that this question is actually asking for the derivative of -sqrt(2x^2-1) evaluated at 1, and write that for your answer.
How does a graph that is continuous around x=1 suddenly become undefined at that very point??? It's not going off to positive or negative inf as it approaches x=1, its very well behaved...
This is just the the negative of the definition of f’(1) where f(x)=sqrt(2x^2-1). This comes from the formula lim as x->a (f(x)-f(a))/(x-a). You get this formula by labeling one point of a graph (a,f(a)), another (x,f(x)) and then computing the slope of the secant line and then taking the limit as x approaches a of the slope to find the instantaneous slope at point (a,f(a)). Using that fact, just take the derivative of sqrt(2x^2-1), multiply it by -1 and then plug in x=1.
Just one question: why don't you use the difference of squares rule also when multiplying the numerator with it's conjugate? It's the same principle as when factoring the difference of squares later on.
So then what did I find when I did direct substitution of the limit of the derivative of the top over the derivative of the bottom, which equals -1? Maybe the idea of convergence is somewhere in there. (It has been a very long time since I learned the basics)
You are right, we couldn't do that for x=1. This is the reason we are using the limit as x approaches 1. Since x isn't 1, but only really close, it isn't 0/0. Isn't that cool? I wish I had commented about that in the video.
Love your videos! Could you explain for the class, when we're doing direct substitution (at time stamp of approximately 0:30), why we don't evaluate the argument of the radical as both the positive and negative versions of the square root of the argument, seeing as how the square root of 1 can either be positive or negative one. When I was teaching math, it was one of those finer details that I always had trouble communicating clearly and concisely. Thanks again for these videos!
@@erin_rose0 While I understand this, I still find it difficult to explain. According to Feynman, this means that I don't understand it well enough. 🤔 This explanation sounds a lot like the, "by definition . . ." explanation that I would often have to use. I, and my students both found this unsatisfying because of all the instances where we would ask them to produce both the positive and negative square root values.
@@jacobmcook The reason for this is that the square root as a function is different from the square root as an inverse operation. Trying to type this made me realize how rare I actually explain this, but when I do, it usually ends with "using the square root to undo an exponent is not the same as looking at sqrt(x) the function".
Yes, for sure L hopital would work great! Current Calc 1 students haven't learned about L hopital yet, so they are solving these by hand using these methods.
@AndyMath As a current Calc 1 student: true and I appreciate having a video explain something we're doing in class, especially as we won't be getting to use L'Hopital on the exam. Edit: first exam. Might learn it later on
Great video but maybe you can consider your target audience for this video because if they clicked on a calculus video they probably know difference of 2 squares formula. I think it would be better for your more complicated videos to skip some easy year 7 explanations. Otherwise cool concept.
Using L'hopital Rule is way more efficient. Take the derivative of the top and bottom. Which gives: lim x-->1 of 0-(4x)(1/2)(2x-1)^-1/2 (Apply chain rule) lim x-->1 of -2x/(2x-1)^1/2 (Make the exponent positive) =-2(1)/(2(1)-1)^1/2 (Substitute 1) =-2/1^1/2 (Simplify) =-2
@@keith6706lim x->1((-2*(x+1))/(1+sqrt(2*x^2-1))) Let x=1 gives ((-2*(1+1))/(1+sqrt(2*1^2-1))) ((-2*(2))/(1+sqrt(2-1))) ((-4)/(1+sqrt(1))) and sqrt(1) is either +1 or -1 (-4/(1+1)) = -2 or (-4/(1-1)) = an infinity
