A small note: At 1:59, the reason why you can exchange the ln and lim operations is because of the continuity of the natural log function ln(x) at x0 = 1 (or, equivalently, because of continuity of the exponential function exp(x) at x0 = 0). It's *not* because of continuity of the function term under the original limit operator.
so this is what I did: take natural log, bring the exponent down rewrote natural log part using property - log(a/b) = log(a) - log(b) rewrote to have the natural log part as numerator, x as denominator (indeterminate, can use L’H) it’s a little tedious, but then it’s very simple, can just use direct substitution then ln(T) = -(ln(a)+ln(b))/2 then do e raised to both sides then simplify to get result.
Let f(x) denote the function term for which we want to determine the limit. Recognizing that the limit is of type 1^inf (or of type 0/0 after taking the natural logarithm), I went for L'Hospital's rule. I first considered the limit of the composite function g(x) = ln(f(x)) as x -> 0. After applying L'Hospital to g(x), I got the limit of -1/2*ln(ab) for g(x) as x -> 0. Utilizing continuity of the exp() function, the limit of f(x) as x -> 0 then becomes 1/sqrt(ab). Pretty easy that way and also a bit more rigorous than using Taylor polynomials as approximations. Speaking of these approximations, I think that PrimeNewtons effectively achieved a similar thing with his Taylor polynomials as what is usually done in the proof of L'Hospital's rule where the functions in the numerator and denominator are replaced by their first-order (i.e. linear) approximations.
Developing a function in a series is a very successful method for calculating limes. I remember that I had an even more complicated limes in my postgraduate exam that could only be solved successfully in this way. What's more, I had to go to the third member of Taylor's series.😎
With only basic knowledge, I managed to solve it in half a page without any approximations, just by using fundamental limits. Still, your way of solving it was interesting!
Solution with no Taylor approximations and no L'H rule: Write (a^x^2+b^x^2)/(a^x+b^x)=1+g(x), where g(x)=(a^x^2-a^x+b^x^2-b^x)/(a^x+b^x): then we need to find the limit of ln(1+g(x))/x. After isolating ln(1+g(x))/g(x) as a factor (which tends to 1), we are left with the limit of g(x)/x, and since the denominator of g(x) tends to 2, we are left with the limit of (a^x^2-a^x+b^x^2-b^x)/x=(a^x^2-a^x)/x+(b^x^2-b^x)/x= =a^x[a^(x^2-x)-1]/x+b^x[b^(x^2-x)-1]/x= =(x-1)a^x[a^(x^2-x)-1]/[x(x-1)]+(x-1)b^x[b^(x^2-x)-1]/[x(x-1)] Since the ratios between the square brackets tend to ln(a) and ln(b), the limit of g(x)/x turns out to be (-lna-lnb)/2=-ln(ab)/2, so that the final answer is (ab)^(-1/2). Also, fun fact: this works for an arbitrary number of parameters a,b,c,...: the limit will be 1/GM(a,b,c,...), where GM denotes the geometric mean.
Sir please do a limit question which was came in JEE Advanced 2014 shift-1 question number 57 it's a question of a limit lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢
The unmatched opening round bracket at 3:45 is never closed; unlike adding zeroes until one reaches the age of 70, which soon becomes a bad choice for getting a lot of zeroes. 😊
I am getting 1 for every value of a and b I put in calculator. And answer is same also by plotting graph. 👉"Limite of a function is the function of the limit" this line is true only when the graph is continuous at that point but ln function is not continuous at x going to zero.
It was possible to use L'Hospital's rule, also known as Bernoulli's rule, and take the derivative of the numerator and denominator of the fraction to reveal the uncertainty of the form 0 divided by 0. Shamanizing with rows is the Stone Age :)
I think one can use L'Hospital's rule for cases when it is 0/0 or infinity/infinity. I don't think it is true here and so L'Hospital's rule cannot be applied. But again, maybe I am missing something.
Using Taylor expansions. T = 1/x * F with F =log((a^x^2 + b^x^2)/(a^x +b^x)). Taylor expansion of F = N/D. Numerator: N= a^x^2+b^x^2=2+x^2*(log(a)+log(b)) + ... Denominator: D= a^x+b^x=2+x(log(a)+log(b))+ … L= log(a)+log(b) . G = (2 + x^2*L)/(2+ x *L). Do a Taylor expansion for G . G(0) = 1 G’ = (2xL(2+xL) - (2+x^2L)L)/( )/(2+ x *L)^2. G’(0) = -L/2. So G = 1 -x*L/2 for small x. Now do a Taylor Expansion for F= log(1 - x*/L/2) ~ -L/2 *x for small x. But T= 1/x * (-L/2*x) = -L/2. So the limit of exp(T) = -exp(log(a)+log(b))/2 = (ab)^(-1/2)
Let’s try to make the long road shorter. f^(1/x) = (1 + f - 1)^(1/x) = (1 + (f - 1))^(1/x) * ((f - 1)/(f - 1)) = (1 + (f - 1))^(1/(f - 1)) * ((f - 1)/x) = [(1 + (f - 1))^(1/(f - 1))]^((f - 1)/x) In our story, x approaches 0, f = f(x) = (a^x^2 + b^x^2)/(a^x + b^x) approaches 1, so (f - 1) approaches 0, and 1/(f - 1) approaches infinity. Therefore, the base of the power function in the in square brackets approaches e. We now need to determine the limit of the exponent ((f - 1)/x) as x approaches 0. ((f - 1)/x) = ((a^x^2 + b^x^2)/(a^x + b^x) - 1) / x = ((a^x^2 + b^x^2) - (a^x + b^x)) / (x * (a^x + b^x)) = ((a^x^2 - a^x)/x + (b^x^2 - b^x)/x) * (1/(a^x + b^x)). The factor 1/(a^x + b^x) approaches ½, so we now need to calculate the limit of (a^x^2 - a^x)/x and similarly (b^x^2 - b^x)/x. This can be done as shown in the video by expanding into a series, or we can use L’Hopital’s rule to obtain the limit: -ln(a) - ln(b), which after substitution gives (-1/2) * ln(ab) = ln((ab)^(1/2)). Finally, e^ln((ab)^(1/2)) = (ab)^(-1/2) = 1/sqrt(ab). You can also try to calculate this limit without series and L'Hopital's rule by extracting a^x from the parentheses and obtaining: (a^x^2 - a^x) / x = (a^x) * (a^(x^2 - x) - 1) / x a^x tends to 1, so we are left with the limit: (a^(x^2 - x) - 1) / x = (e^((x^2 - x) * ln(a)) - 1) / x as x tends to 0 Here, you can use the expansion in a series. After subtracting 1, each term can be divided by x, and we see that all terms except for -ln(a) tend to 0. You can continue further: (e^((x^2 - x) * ln(a)) - 1) / x = (e^((x^2 - x) * ln(a)) - 1) / ((x^2 - x) * ln(a)) * ((x^2 - x) * ln(a)) / x = ((e^t - 1) / t) * (x - 1) * ln(a) As you can see, the remaining proof is that (e^t - 1) / t tends to 1 as t tends to 0. This identity is one of the fundamental limits in calculus
Take logs, and you want lim as x goes to zero of 1/x * (ln(a^(x^2) + b^(x^2)) - ln(a^x + b^x) ), which equals the sum of the limits of 1/x * (ln(a^(x^2) + b^(x^2)) - 2) and -1/x * (ln(a^x + b^x) - 2). The first is the derivative of ln(a^(x^2) + b^(x^2)) at x = 0 and the second is -1 times the derivative of ln(a^x + b^x) at x = 0. The first derivative computes to zero and the second derivative to (ln(a) + ln (b))/2 = ln(ab)/2. So the log of the limit is -ln(ab)/2 and thus the limit is 1/(ab)^(1/2).
My experience with LaTeX is very limited. Maybe in the future, when I really learn it well. Also, I've not been as active as I used to be. Hope things change soon.
@@PrimeNewtons So how you write documents about your researches on open problems or at least how you prepare scripts and other documents for students ?
Hey there people, on a very unrelated note I need help at proving e^x>x^e unless x=e , I had read this recently but unfortunately the proof of it wasn't included in the book it was only stated as a fact
The Taylor series is an approximation of the function near a specific point. As long as you keep enough terms, so that the function is approximated accurately, the taylor series can always be used in order to calculate a limit. my point is that there is not much point in using something that is much more advanced in order calculate something much simpler. De l' hopital rule is essentially the same thing. Given two functions f(x) and g(x), one uses the linear approximations f(x)=f(x_0)+f'(x_0)(x-x_0) and g(x)=g(x_0)+g'(x_0)(x-x_0). Obviously if f(x_0)=g(x_0)= 0 the ratio f(x)/g(x) around x_0 is f'(x_0)/g'(x_0) .
@@stealth3122 I don't see his email anywhere during the video. Just tell me what his email address is. I don't do twitter, facebook, other social media.
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You’re a happy person.
An absolute banger of a limit. That I wish I never encounter in an exam.
The last question: 👁👄👁
@heinrich.hitzinger I wont sleep tonight out of pure dread
That was one heck of a limit! I was expecting you to use L'Hospital's Rule. But your method was definitely more intriguing and challenging!
Does L'hopital work for (1)^1/0 ?
@@shhh._ You can take the log and manipulate it to arrange it as 0/0.
A small note: At 1:59, the reason why you can exchange the ln and lim operations is because of the continuity of the natural log function ln(x) at x0 = 1 (or, equivalently, because of continuity of the exponential function exp(x) at x0 = 0). It's *not* because of continuity of the function term under the original limit operator.
I love your solution! When I solved it I just used L'H and called it a day, but your insights make this so much more interesting!
I guess it is a general rule: lim(x->0) of (f(x))^(1/x), where f(0)=1 and f'(x) is continuous around f'(0) . The limit is e^(f'(x)).
so this is what I did:
take natural log, bring the exponent down
rewrote natural log part using property - log(a/b) = log(a) - log(b)
rewrote to have the natural log part as numerator, x as denominator (indeterminate, can use L’H)
it’s a little tedious, but then it’s very simple, can just use direct substitution
then ln(T) = -(ln(a)+ln(b))/2
then do e raised to both sides
then simplify to get result.
Let f(x) denote the function term for which we want to determine the limit. Recognizing that the limit is of type 1^inf (or of type 0/0 after taking the natural logarithm), I went for L'Hospital's rule. I first considered the limit of the composite function g(x) = ln(f(x)) as x -> 0. After applying L'Hospital to g(x), I got the limit of -1/2*ln(ab) for g(x) as x -> 0. Utilizing continuity of the exp() function, the limit of f(x) as x -> 0 then becomes 1/sqrt(ab).
Pretty easy that way and also a bit more rigorous than using Taylor polynomials as approximations. Speaking of these approximations, I think that PrimeNewtons effectively achieved a similar thing with his Taylor polynomials as what is usually done in the proof of L'Hospital's rule where the functions in the numerator and denominator are replaced by their first-order (i.e. linear) approximations.
Fantastic. I wish I forget this video so I could try to do this limit by myself
Wowza! This question requires a lot of brains!😊
Thank you for this! You’re the best
Developing a function in a series is a very successful method for calculating limes. I remember that I had an even more complicated limes in my postgraduate exam that could only be solved successfully in this way. What's more, I had to go to the third member of Taylor's series.😎
With only basic knowledge, I managed to solve it in half a page without any approximations, just by using fundamental limits. Still, your way of solving it was interesting!
Really needed this one 😊
I solved it.
After solving I felt happiness.
Solution with no Taylor approximations and no L'H rule:
Write (a^x^2+b^x^2)/(a^x+b^x)=1+g(x), where g(x)=(a^x^2-a^x+b^x^2-b^x)/(a^x+b^x): then we need to find the limit of ln(1+g(x))/x.
After isolating ln(1+g(x))/g(x) as a factor (which tends to 1), we are left with the limit of g(x)/x, and since the denominator of g(x) tends to 2, we are left with the limit of
(a^x^2-a^x+b^x^2-b^x)/x=(a^x^2-a^x)/x+(b^x^2-b^x)/x=
=a^x[a^(x^2-x)-1]/x+b^x[b^(x^2-x)-1]/x=
=(x-1)a^x[a^(x^2-x)-1]/[x(x-1)]+(x-1)b^x[b^(x^2-x)-1]/[x(x-1)]
Since the ratios between the square brackets tend to ln(a) and ln(b), the limit of g(x)/x turns out to be (-lna-lnb)/2=-ln(ab)/2, so that the final answer is (ab)^(-1/2).
Also, fun fact: this works for an arbitrary number of parameters a,b,c,...: the limit will be 1/GM(a,b,c,...), where GM denotes the geometric mean.
Sir please do a limit question which was came in
JEE Advanced 2014 shift-1 question number 57
it's a question of a limit
lim as x-->1
[{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4
You have to find the greatest value of a
It has 2 possible answers 0 and 2
But I want the reason that why should I reject 2 and accept 0
Because final answer is 0
Please help 😢
Ok. I've seen this request before. Ill check it out. Please email me if you have the full question
@PrimeNewtons please tell your Email ID
@@PrimeNewtons sir please tell your E mail ID
@@PrimeNewtons Sir please tell your E-m@il ID
The unmatched opening round bracket at 3:45 is never closed; unlike adding zeroes until one reaches the age of 70, which soon becomes a bad choice for getting a lot of zeroes. 😊
Wow!
Use the formula for 1^infinity, and then L', it only takes 6 steps (5mins).
I would either do L’Hopital’s Rule or a full series expansion.
That's crazy 💀
You really need to do the full series expansion to be rigorous.
I am getting 1 for every value of a and b I put in calculator. And answer is same also by plotting graph.
👉"Limite of a function is the function of the limit" this line is true only when the graph is continuous at that point but ln function is not continuous at x going to zero.
It was possible to use L'Hospital's rule, also known as Bernoulli's rule, and take the derivative of the numerator and denominator of the fraction to reveal the uncertainty of the form 0 divided by 0. Shamanizing with rows is the Stone Age :)
I think one can use L'Hospital's rule for cases when it is 0/0 or infinity/infinity. I don't think it is true here and so L'Hospital's rule cannot be applied.
But again, maybe I am missing something.
Using Taylor expansions.
T = 1/x * F with F =log((a^x^2 + b^x^2)/(a^x +b^x)). Taylor expansion of F = N/D. Numerator: N= a^x^2+b^x^2=2+x^2*(log(a)+log(b)) + ... Denominator: D= a^x+b^x=2+x(log(a)+log(b))+ … L= log(a)+log(b) . G = (2 + x^2*L)/(2+ x *L). Do a Taylor expansion for G . G(0) = 1 G’ = (2xL(2+xL) - (2+x^2L)L)/( )/(2+ x *L)^2. G’(0) = -L/2. So G = 1 -x*L/2 for small x. Now do a Taylor Expansion for F= log(1 - x*/L/2) ~ -L/2 *x for small x. But T= 1/x * (-L/2*x) = -L/2. So the limit of exp(T) = -exp(log(a)+log(b))/2 = (ab)^(-1/2)
What about a substitution c = b/a? We have a,b>0 so c definetely exists. Then we have the two second wonderful limits. At least something like that.
Let’s try to make the long road shorter.
f^(1/x) = (1 + f - 1)^(1/x) = (1 + (f - 1))^(1/x) * ((f - 1)/(f - 1)) = (1 + (f - 1))^(1/(f - 1)) * ((f - 1)/x) = [(1 + (f - 1))^(1/(f - 1))]^((f - 1)/x)
In our story, x approaches 0, f = f(x) = (a^x^2 + b^x^2)/(a^x + b^x) approaches 1, so (f - 1) approaches 0, and 1/(f - 1) approaches infinity. Therefore, the base of the power function in the in square brackets approaches e.
We now need to determine the limit of the exponent ((f - 1)/x) as x approaches 0.
((f - 1)/x) = ((a^x^2 + b^x^2)/(a^x + b^x) - 1) / x = ((a^x^2 + b^x^2) - (a^x + b^x)) / (x * (a^x + b^x)) = ((a^x^2 - a^x)/x + (b^x^2 - b^x)/x) * (1/(a^x + b^x)).
The factor 1/(a^x + b^x) approaches ½, so we now need to calculate the limit of (a^x^2 - a^x)/x and similarly (b^x^2 - b^x)/x.
This can be done as shown in the video by expanding into a series, or we can use L’Hopital’s rule to obtain the limit:
-ln(a) - ln(b), which after substitution gives (-1/2) * ln(ab) = ln((ab)^(1/2)).
Finally, e^ln((ab)^(1/2)) = (ab)^(-1/2) = 1/sqrt(ab).
You can also try to calculate this limit without series and L'Hopital's rule by extracting a^x from the parentheses and obtaining:
(a^x^2 - a^x) / x = (a^x) * (a^(x^2 - x) - 1) / x
a^x tends to 1, so we are left with the limit:
(a^(x^2 - x) - 1) / x = (e^((x^2 - x) * ln(a)) - 1) / x as x tends to 0
Here, you can use the expansion in a series. After subtracting 1, each term can be divided by x, and we see that all terms except for -ln(a) tend to 0.
You can continue further:
(e^((x^2 - x) * ln(a)) - 1) / x = (e^((x^2 - x) * ln(a)) - 1) / ((x^2 - x) * ln(a)) * ((x^2 - x) * ln(a)) / x = ((e^t - 1) / t) * (x - 1) * ln(a)
As you can see, the remaining proof is that (e^t - 1) / t tends to 1 as t tends to 0. This identity is one of the fundamental limits in calculus
Take logs, and you want lim as x goes to zero of 1/x * (ln(a^(x^2) + b^(x^2)) - ln(a^x + b^x) ), which equals the sum of the limits of
1/x * (ln(a^(x^2) + b^(x^2)) - 2) and -1/x * (ln(a^x + b^x) - 2). The first is the derivative of ln(a^(x^2) + b^(x^2)) at x = 0 and the second is -1 times the derivative of ln(a^x + b^x) at x = 0. The first derivative computes to zero and the second derivative to (ln(a) + ln (b))/2 = ln(ab)/2. So the log of the limit is -ln(ab)/2 and thus the limit is 1/(ab)^(1/2).
Can you untegrate fraction part and absolute function?!
You’re hezekiah’s dad! I go to the same school as him
Actually, it's a "VERY SHORT" road.
One might say this really pushed you to your limits.
You have your algebra series
Can you record series about LaTeX for viewers who want to make pdf based on your algebra series
My experience with LaTeX is very limited. Maybe in the future, when I really learn it well. Also, I've not been as active as I used to be. Hope things change soon.
@@PrimeNewtons So how you write documents about your researches on open problems
or at least how you prepare scripts and other documents for students ?
Instead of using approximations, you could have invoked Taylor's theorem, and just had a bound or you could have used o and O notation.
T=Limit[(a*x^2+b*x^2)/(a^x+b^x),x->0]^(1/x),a,b>0 T=1/Sqrt[ab]=Sqrt[ab]/ab It’s in my head.
I have asked this before but where did you get your math education and do you teach somewhere?
As a practical matter, using L'Hôpital's Rule is much quicker. As a teaching tool, the methods you used are far more interesting.
Is the limit the reason that the domains of a and b are restricted?
Hey there people, on a very unrelated note I need help at proving e^x>x^e unless x=e , I had read this recently but unfortunately the proof of it wasn't included in the book it was only stated as a fact
just use de l' hopital rule. It doesn't make much sense using taylor expansions to calculate a limit.
It's amazing that it can be done that way! (with the Taylor series)
The Taylor series is an approximation of the function near a specific point. As long as you keep enough terms, so that the function is approximated accurately, the taylor series can always be used in order to calculate a limit. my point is that there is not much point in using something that is much more advanced in order calculate something much simpler.
De l' hopital rule is essentially the same thing. Given two functions f(x) and g(x), one uses the linear approximations f(x)=f(x_0)+f'(x_0)(x-x_0) and g(x)=g(x_0)+g'(x_0)(x-x_0). Obviously if f(x_0)=g(x_0)= 0 the ratio f(x)/g(x) around x_0 is f'(x_0)/g'(x_0) .
Sir please tell your Email ID
wdym? his email is at the start of the video always.
@@stealth3122 I don't see his email anywhere during the video. Just tell me what his email address is. I don't do twitter, facebook, other social media.