For those that are confused near the end. When he says 'take the dominating factor', he is NOT using L'H. He is multiplying the numerator and denominator by 1/x and then taking the limit in one step Edit: If the dominating factor was x^2 then he would have multiplied the numerator and denominator by 1/ (x^2)
@@derekschmidt5705 The second last limit has the function f(x) = 4x/[√(x² + 4x) + x]. You can't just substitute x by infinity on this function because then it gives the infinity/infinity indetermination. Taking the "dominating factor" is not magic. Here's what it is: x approaches infinity, so x is just a big positive number. That implies x ≠ 0. Hence, we can just multiply the function by (1/x)/(1/x), which is equivalent to multiply by 1 and does not change anything but the expression of the function: 4x/[√(x² + 4x) + x] = {4x/[√(x² + 4x) + x]}[(1/x)/(1/x)] = 4/[(1/x)√(x² + 4x) + 1] Multiplying a square root by (1/x), where x > 0 because x approaches infinity, is the same as taking just the square root, but multiplying its interior by (1/x)² {since √[(1/x)²] = 1/x}. Therefore, 4x/[√(x² + 4x) + x] = 4/{√[(1/x)²(x² + 4x)] + 1} = 4/[√(1 + 4/x) + 1] Notice that 4/x approaches 0 when x approaches infinity, so the limit is L = 4/[√(1 + 0) + 1] = 4/[√(1) + 1] = 4/(1 + 1) = 4/2 = 2 as is stated in the video.
@@diegocabrales That is the safe method, but what he did in the video is just "x + x, we get 2x on the bottom". Am I missing something here? What steps did he skip to come to this conclusion that just adding those two x's will give us 2x on the bottom?
I never trust myself to get the "dominant part" step right; my judgment is poor. So at that point I would divide top and bottom by x, which would force almost all the x's to go away, except for the "4x" under the radical which would become "4/x". And I am very confident that term will go to 0 as x goes to infinity. Anyway, that's how clods like me do it safely.
I don't think education surrounding functions is imbued with the right level of respect. We need to make sure various "properties", such as f(x)=f(x+h)-f(h), aren't naturally assumed to be true.
So taking the dominating factor under the radical means disregarding the 4X under the radical and just taking the sqrt of X squared! Which is X. Then add this X with the other X which gives 2X in the denominator. The 4X in the numerator is unchanged so 4X/2X = 2. Easy.
This is the ideal and recommended way of doing this. But another possible way is to turn ∞-∞ form to 0/0 form by putting (t=1/x) so as x tends to ∞, t tends to zero, and after simplifying, your limit becomes ((√(1+4t)) -1)/t. Now we are able to use l`hôpital which reduces it to (2/√(1+4t)) which on putting t=0 gives 2.
another solution: take x² common from x²+4x, and take this x out of the sqrt so it becomes: x*sqrt(1+4/x) - x which simplifies to : x*(sqrt(1+4/x)-1), which can be written as x*( (1+4/x)^(0.5) - 1) as x gets really big, 4/x gets really small, hence you can use binomial approximation (which will be an exact value instead of approximation at infinity): x*( 1 + 4/x*0.5 - 1) = x*(2/x) = 2
To be fair, there is some benefit to doing the sqrt((x + 2)^2 - 4) thing. We want lim f(x) as x -> inf, where f(x) = sqrt(x^2 + 4x) - x f(x) = sqrt(x^2 + 4x) - x f(x) = sqrt(x^2 + 4x + 4 - 4) - x f(x) < sqrt(x^2 + 4x + 4) - x f(x) < x + 2 - x f(x) < 2 Thus, the limit will be less than or equal to 2.
We can just say, that - because x goes to infinity - a fixed extra term of +4 is so insignificant, that it doesn't change the limit. As such, √(x² + 4x) ≅ √(x² + 4x + 4) = √(x + 2)² = x + 2 So the limit term is essentially x + 2 - x, which is a constant 2.
@@chaosredefined3834 as long as the term is dependent on x - which is the case here, as it is part of the square - it should always work. Of course, if you add a term outside the square, or if the square didn't have an x, then it obviously wont work.
In this case you can add the four inside of the radical and not mess with the other side but that requires calculus to prove but the propogated error merges to zero
You can still solve this by completing the square . Just add 4 and subtract 4 inside the radical. Then you get the lim√((x+2)²-4) -x = lim(x + 2 - x) = 2. The -4 is negligible in the limit to infinity.
I think the OP was confused about why its not allowed to say √(x) is approximately equal to √(x + 4) - √(4) for large x. To be fair I'm not sure how I would answer this either
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I'd say that for large x, sqrt(x+4) will approach sqrt(x), but sqrt(4) will not approach 0.
When you said it's not helpful, maybe it can in your example. You can factorize the difference of two square and then get two square roots. But maybe it's not helpful anyway...
In this case, you can also just say, that - since x goes to infinity - a +4 in the root is insignificant. Therefore the Term can be just assumed to be √(x² + 4x + 4) - x = √(x + 2)² - x = x + 2 - x = 2 So much simpler than the solution in the video.
Sadly this has become typical. I have a calc 3 student who still seems to be struggling with fractions, and I quite often see much worse. Been college math tutoring for well over 7 years
The limit of the actual denominator is equally complicated to calculate as the original fraction, so you will essentially just go in circles using L'Hospital here
Your denominator is wrong. If you differentiate the denomintor you should get 1/[2sqrt(x^2+4x)] * (2x+4) + 1 =(2x+4)/[2sqrt(x^2+4x)] + 1 =(x+2)/sqrt(x^2+4x) + 1 And if you put that on the denominator the expression would be more complicated than the original which is counter productive
Why even make it so difficult? We are calculating a limit towards infinity. So adding a constant term is basically meaningless. So as such, we can just add a +4 inside the square to get √(x² + 4x + 4), which resolves to x + 2. This means the original term is now x + 2 - x, which is just 2, which is the correct limit. There is absolutely no need for complicated algorithms like L'Hospital or even multiply by its conjugate.
Check out more similar calculus tutorials: th-cam.com/video/_ZCijObsrHQ/w-d-xo.html
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For those that are confused near the end. When he says 'take the dominating factor', he is NOT using L'H. He is multiplying the numerator and denominator by 1/x and then taking the limit in one step
Edit: If the dominating factor was x^2 then he would have multiplied the numerator and denominator by 1/ (x^2)
I feel like I understood the video just fine, and you just confused me.
@@derekschmidt5705 The second last limit has the function f(x) = 4x/[√(x² + 4x) + x].
You can't just substitute x by infinity on this function because then it gives the infinity/infinity indetermination.
Taking the "dominating factor" is not magic. Here's what it is:
x approaches infinity, so x is just a big positive number. That implies x ≠ 0. Hence, we can just multiply the function by (1/x)/(1/x), which is equivalent to multiply by 1 and does not change anything but the expression of the function:
4x/[√(x² + 4x) + x] = {4x/[√(x² + 4x) + x]}[(1/x)/(1/x)] = 4/[(1/x)√(x² + 4x) + 1]
Multiplying a square root by (1/x), where x > 0 because x approaches infinity, is the same as taking just the square root, but multiplying its interior by (1/x)² {since √[(1/x)²] = 1/x}. Therefore,
4x/[√(x² + 4x) + x] = 4/{√[(1/x)²(x² + 4x)] + 1} = 4/[√(1 + 4/x) + 1]
Notice that 4/x approaches 0 when x approaches infinity, so the limit is
L = 4/[√(1 + 0) + 1] = 4/[√(1) + 1] = 4/(1 + 1) = 4/2 = 2
as is stated in the video.
@@diegocabrales That is the safe method, but what he did in the video is just "x + x, we get 2x on the bottom". Am I missing something here? What steps did he skip to come to this conclusion that just adding those two x's will give us 2x on the bottom?
I never trust myself to get the "dominant part" step right; my judgment is poor. So at that point I would divide top and bottom by x, which would force almost all the x's to go away, except for the "4x" under the radical which would become "4/x". And I am very confident that term will go to 0 as x goes to infinity. Anyway, that's how clods like me do it safely.
I don't think education surrounding functions is imbued with the right level of respect. We need to make sure various "properties", such as f(x)=f(x+h)-f(h), aren't naturally assumed to be true.
Agreed
So taking the dominating factor under the radical means disregarding the 4X under the radical and just taking the sqrt of X squared! Which is X. Then add this X with the other X which gives 2X in the denominator. The 4X in the numerator is unchanged so 4X/2X = 2. Easy.
Thanks. This made me understand it
This is the ideal and recommended way of doing this. But another possible way is to turn ∞-∞ form to 0/0 form by putting (t=1/x) so as x tends to ∞, t tends to zero, and after simplifying, your limit becomes ((√(1+4t)) -1)/t. Now we are able to use l`hôpital which reduces it to (2/√(1+4t)) which on putting t=0 gives 2.
another solution:
take x² common from x²+4x, and take this x out of the sqrt
so it becomes: x*sqrt(1+4/x) - x
which simplifies to : x*(sqrt(1+4/x)-1), which can be written as x*( (1+4/x)^(0.5) - 1)
as x gets really big, 4/x gets really small, hence you can use binomial approximation (which will be an exact value instead of approximation at infinity):
x*( 1 + 4/x*0.5 - 1) = x*(2/x) = 2
I have thought the same method to solve this limit.
To be fair, there is some benefit to doing the sqrt((x + 2)^2 - 4) thing.
We want lim f(x) as x -> inf, where f(x) = sqrt(x^2 + 4x) - x
f(x) = sqrt(x^2 + 4x) - x
f(x) = sqrt(x^2 + 4x + 4 - 4) - x
f(x) < sqrt(x^2 + 4x + 4) - x
f(x) < x + 2 - x
f(x) < 2
Thus, the limit will be less than or equal to 2.
We can just say, that - because x goes to infinity - a fixed extra term of +4 is so insignificant, that it doesn't change the limit.
As such, √(x² + 4x) ≅ √(x² + 4x + 4) = √(x + 2)² = x + 2
So the limit term is essentially x + 2 - x, which is a constant 2.
@@m.h.6470 Need to be careful doing that. It works here, but... proceed with caution.
@@chaosredefined3834 as long as the term is dependent on x - which is the case here, as it is part of the square - it should always work. Of course, if you add a term outside the square, or if the square didn't have an x, then it obviously wont work.
In this case you can add the four inside of the radical and not mess with the other side but that requires calculus to prove but the propogated error merges to zero
5:00 i think a safer step would be to divide up & down by x and get 4/(sqrt(1+4/x) +1) that tends to 2.
You can still solve this by completing the square . Just add 4 and subtract 4 inside the radical. Then you get the lim√((x+2)²-4) -x = lim(x + 2 - x) = 2. The -4 is negligible in the limit to infinity.
I think the OP was confused about why its not allowed to say √(x) is approximately equal to √(x + 4) - √(4) for large x. To be fair I'm not sure how I would answer this either
I'd say that for large x, sqrt(x+4) will approach sqrt(x), but sqrt(4) will not approach 0.
Ah right, that makes a lot of sense
Rationalise and divde by x we get ans as 4/2 i.e 2 :)
Totally unbelievable!!
Why did you not use the 4x at the end ? Wouldn’t that have changed the answer
When you said it's not helpful, maybe it can in your example. You can factorize the difference of two square and then get two square roots. But maybe it's not helpful anyway...
4/(sqrt(1+4/x)+1)=4/(sqrt(1+0)+1)=4/(1+1)=4/2=2
In this case, you can also just say, that - since x goes to infinity - a +4 in the root is insignificant.
Therefore the Term can be just assumed to be √(x² + 4x + 4) - x = √(x + 2)² - x = x + 2 - x = 2
So much simpler than the solution in the video.
5:22 factor x from the denominator
just complete the square easiest limit ever
Bros doing calc and still messing this up
Sadly this has become typical. I have a calc 3 student who still seems to be struggling with fractions, and I quite often see much worse. Been college math tutoring for well over 7 years
Whoever asked this should not be doing calculus, he/she should go back to 5th grade
I don't get the end.
start from lim x→∞ (4x)/(√(x²+4x)+x), use l`Hospital, 4/((2x+4)/(2√x))+1=4/∞=0
I don’t think i would use L’Hospital for this limit tho, i would take out an x at numerator and denominator, resulting in 1/2 like in the video
You made a mistake when differentiating the denominator.
f(x)=sqrt(x²+4x)+x
Thus
f'(x)=(x+2)/sqrt(x²+4x) + 1
The limit of the actual denominator is equally complicated to calculate as the original fraction, so you will essentially just go in circles using L'Hospital here
Your denominator is wrong.
If you differentiate the denomintor you should get
1/[2sqrt(x^2+4x)] * (2x+4) + 1
=(2x+4)/[2sqrt(x^2+4x)] + 1
=(x+2)/sqrt(x^2+4x) + 1
And if you put that on the denominator the expression would be more complicated than the original which is counter productive
Why even make it so difficult? We are calculating a limit towards infinity. So adding a constant term is basically meaningless.
So as such, we can just add a +4 inside the square to get √(x² + 4x + 4), which resolves to x + 2.
This means the original term is now x + 2 - x, which is just 2, which is the correct limit.
There is absolutely no need for complicated algorithms like L'Hospital or even multiply by its conjugate.