In this video , I showed how to compute the limit of all absolute value functions
Пікірлер: 37
@aavalos77605 ай бұрын
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
@shmuelzehavi49405 ай бұрын
That's right. I used the same approach.
@JaydenPatrick-jy3mr3 ай бұрын
fr the most underrated education youtuber
@fsponj2 ай бұрын
Yeah
@vitotozzi19725 ай бұрын
Newtons, your explain is really excelent....
@PrimeNewtons5 ай бұрын
Thank you kindly!
@evgeniospagkalis9922Ай бұрын
Great video!!
@subbaraooruganti11 күн бұрын
Excellent explanation
@michaelbaum67965 ай бұрын
Excellent explanation- great👌
@manojitmaity78935 ай бұрын
You are just amazing!!
@pratapray426418 күн бұрын
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.
@surendrakverma5555 ай бұрын
Very good. Thanks Sir
@kingbeauregard5 ай бұрын
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.
@mattbrown5125 ай бұрын
This function has such a cool-looking graph.
@flowingafterglow6295 ай бұрын
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.
@MalosePeterMogodi5 ай бұрын
the pause at 4:28 killed me sir😅
@MalosePeterMogodi5 ай бұрын
sorry 4:23+
@Esraa-pf5dg25 күн бұрын
جميل جدا
@tomasbeltran040505 ай бұрын
Yesss I got it
@klementhajrullaj12225 ай бұрын
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. 😀😉
@GreenMeansGOF5 ай бұрын
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.
@AubreyForever5 ай бұрын
Great!
@Gleb7245 ай бұрын
I expanded the fraction by modulo x and diveded numeretor and denominator and subtituted x equal to zero.
@Gleb7245 ай бұрын
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.
@d.yousefsobh70105 ай бұрын
Very good
@user-gu6dc7yu1m5 ай бұрын
Two conditions: x is positive or negative!
@easymaths_4u5 ай бұрын
Why did he not mentioned abs(x) for x=0 ,for x≥0 ,|x|=x and for x
@jumpman82825 ай бұрын
@@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.
@rimantasri45785 ай бұрын
9:52 but if you're looking for x > 0 then not only 0
@williamperez-hernandez39685 ай бұрын
Taking the domain x>1 does not let you take the correct limit x approaching zero.
@77Chester775 ай бұрын
Bravo
@michelmegabacus78945 ай бұрын
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
@klementhajrullaj12225 ай бұрын
And if, |x^3-x|/(|x^3|-|x|), or |x^3-x|/(|x^3|-x)??? ...
@user-kk6rj8zn6p2 ай бұрын
I have heard do not enter 😂😂
@JSSTyger5 ай бұрын
I'll say the limit is -1.
@JSSTyger5 ай бұрын
Yesss I'm right. I'm 42 and never stopped learning. **flexes**
@samar59925 ай бұрын
No need of simplifying so much, Left hand Limit = -1 Right Hand limit = -1 Hence limit is -1