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.
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.
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. 😀😉
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
Newtons, your explain is really excelent....
Thank you kindly!
fr the most underrated education youtuber
Yeah
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 👍
amazing explanation,thank you so much.
Excellent explanation- great👌
You are just amazing!!
Thanks you sir its a good one
Very good. Thanks Sir
Excellent explanation
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.
Great video!!
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.
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.
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.
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.
the pause at 4:28 killed me sir😅
sorry 4:23+
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. 😀😉
Very good
جميل جدا
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.
Great!
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.
Bravo
Yesss I got it
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
And if, |x^3-x|/(|x^3|-|x|), or |x^3-x|/(|x^3|-x)??? ...
I'll say the limit is -1.
Yesss I'm right. I'm 42 and never stopped learning. **flexes**
I have heard do not enter 😂😂
No need of simplifying so much,
Left hand Limit = -1
Right Hand limit = -1
Hence limit is -1