OMG finally someone explains this concept clearly! I had a problem in my calc syllabus that was driving me insane and finally solved it thanks to your video; tysm!!
We can do this without worrying about approaching from left or right actually. The key is realizing x^3 = x* x^2 = x*|x||x|. f(x) = |x^3 - x| / (x^3 - |x|) = (|x|*|x^2 - 1|) / (x*|x||x| - |x|) = |x^2 - 1| / (x|x| - 1) Taking the limit to 0 gives us |-1| / (-1) = 1/(-1) = -1
Yeah, I had to plot it out just to see it. For x>1, the answer is 1. For 0 < x < 1, the answer is -1. From -1 to 0, it's almost linear from 0 to -1, and then for x < -1, it starts at 0 and asymptotes to -1. Really weird.
I found an easier way. First divide both numerator and denominator by |x|. Then the numerator becomes |x^2-1|. The denominator becomes x^3/|x|-1 which simplifies to x|x|-1. Thus we have the limit of |x^2-1|/(x|x|-1). Plug in 0 and we get |-1|/(-1)=-1.
Beauty limits are even: a) V(x-1)/(Vx-1) when x goes to 1 b) (8^x-1)/(4^x-1) when x goes to 0 c) (4^x-2^x)/(2^x-1) when x goes to 0 d) [log with base 2 of (x-1)]/[(log with base 2 of x)-1] when x goes to 2. 😀😉
Exlaination |x³-x|/(x³-|x|)=|x||x²-1|/(x|x|²-|x|),divide nominator and denominator by |x| and we have. |x²-1|/(x|x|-1) and subtitute x=0 and we have |0-1|/(0-1)=1/(-1)=-1 and it's a right answer.
Autre solution. Au voisinage de 0, un polynôme est équivalent à son terme non nul de plus bas degré. Si x>0, au voisinage de 0, x³ - |x| = x³ - x ~ -x = -|x| Si x
fr the most underrated education youtuber
Yeah
OMG finally someone explains this concept clearly! I had a problem in my calc syllabus that was driving me insane and finally solved it thanks to your video; tysm!!
I love the way you teach. I'll make sure to imitate you when I'm doing my tutoring job!
We can do this without worrying about approaching from left or right actually. The key is realizing x^3 = x* x^2 = x*|x||x|.
f(x) = |x^3 - x| / (x^3 - |x|)
= (|x|*|x^2 - 1|) / (x*|x||x| - |x|)
= |x^2 - 1| / (x|x| - 1)
Taking the limit to 0 gives us |-1| / (-1) = 1/(-1) = -1
That's right. I used the same approach.
This is a gorgeous approach, and it deserves a thumbs-up 👍
Newtons, your explain is really excelent....
Thank you kindly!
This function has such a cool-looking graph.
Yeah, I had to plot it out just to see it. For x>1, the answer is 1. For 0 < x < 1, the answer is -1. From -1 to 0, it's almost linear from 0 to -1, and then for x < -1, it starts at 0 and asymptotes to -1.
Really weird.
Excellent explanation- great👌
amazing explanation,thank you so much.
You are just amazing!!
love ur channel
Excellent explanation
Very good. Thanks Sir
I found an easier way. First divide both numerator and denominator by |x|. Then the numerator becomes |x^2-1|. The denominator becomes
x^3/|x|-1
which simplifies to x|x|-1. Thus we have the limit of
|x^2-1|/(x|x|-1).
Plug in 0 and we get |-1|/(-1)=-1.
Thanks you sir its a good one
I would've gotten this wrong, at least on the first pass. I wouldn't have considered that the range from -1 to +1 behaves differently.
the pause at 4:28 killed me sir😅
sorry 4:23+
Great video!!
Use |x^3-x| = |x.(x^2-1)| =|x|.|x^2-1| and in the denominator x^3-|x| = |x| .( |x|.x-1) , it will be more simpler to prove.
Beauty limits are even: a) V(x-1)/(Vx-1) when x goes to 1 b) (8^x-1)/(4^x-1) when x goes to 0 c) (4^x-2^x)/(2^x-1) when x goes to 0 d) [log with base 2 of (x-1)]/[(log with base 2 of x)-1] when x goes to 2. 😀😉
I expanded the fraction by modulo x and diveded numeretor and denominator and subtituted x equal to zero.
Exlaination |x³-x|/(x³-|x|)=|x||x²-1|/(x|x|²-|x|),divide nominator and denominator by |x| and we have. |x²-1|/(x|x|-1) and subtitute x=0 and we have |0-1|/(0-1)=1/(-1)=-1 and it's a right answer.
Very good
Yesss I got it
Great!
جميل جدا
Bravo
Two conditions: x is positive or negative!
Why did he not mentioned abs(x) for x=0 ,for x≥0 ,|x|=x and for x
@@easymaths_4u He omitted 𝑥 = 0 and 𝑥 = 1 because the function that we are taking the limit of is not defined for those values.
Autre solution.
Au voisinage de 0, un polynôme est équivalent à son terme non nul de plus bas degré.
Si x>0, au voisinage de 0, x³ - |x| = x³ - x ~ -x = -|x|
Si x
9:52 but if you're looking for x > 0 then not only 0
Taking the domain x>1 does not let you take the correct limit x approaching zero.
And if, |x^3-x|/(|x^3|-|x|), or |x^3-x|/(|x^3|-x)??? ...
I have heard do not enter 😂😂
I'll say the limit is -1.
Yesss I'm right. I'm 42 and never stopped learning. **flexes**
No need of simplifying so much,
Left hand Limit = -1
Right Hand limit = -1
Hence limit is -1