Limit of absolute value functions [ lim |x^3 - x|/(x^3

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Limit of absolute value functions [ lim |x^3 - x|/(x^3 - |x|) as x goes to 0 ]

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Prime Newtons

5 ай бұрын

In this video , I showed how to compute the limit of all absolute value functions

Пікірлер: 37

@aavalos7760 5 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

@shmuelzehavi4940 5 ай бұрын

That's right. I used the same approach.

@JaydenPatrick-jy3mr 3 ай бұрын

fr the most underrated education youtuber

@fsponj 2 ай бұрын

Yeah

@vitotozzi1972 5 ай бұрын

Newtons, your explain is really excelent....

@PrimeNewtons 5 ай бұрын

Thank you kindly!

@evgeniospagkalis9922 Ай бұрын

Great video!!

@subbaraooruganti 11 күн бұрын

Excellent explanation

@michaelbaum6796 5 ай бұрын

Excellent explanation- great👌

@manojitmaity7893 5 ай бұрын

You are just amazing!!

@pratapray4264 18 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.

@surendrakverma555 5 ай бұрын

Very good. Thanks Sir

@kingbeauregard 5 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.

@mattbrown512 5 ай бұрын

This function has such a cool-looking graph.

@flowingafterglow629 5 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.

@MalosePeterMogodi 5 ай бұрын

the pause at 4:28 killed me sir😅

@MalosePeterMogodi 5 ай бұрын

sorry 4:23+

@Esraa-pf5dg 25 күн бұрын

جميل جدا

@tomasbeltran04050 5 ай бұрын

Yesss I got it

@klementhajrullaj1222 5 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. 😀😉

@GreenMeansGOF 5 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.

@AubreyForever 5 ай бұрын

Great!

@Gleb724 5 ай бұрын

I expanded the fraction by modulo x and diveded numeretor and denominator and subtituted x equal to zero.

@Gleb724 5 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.yousefsobh7010 5 ай бұрын

Very good

@user-gu6dc7yu1m 5 ай бұрын

Two conditions: x is positive or negative!

@easymaths_4u 5 ай бұрын

Why did he not mentioned abs(x) for x=0 ,for x≥0 ,|x|=x and for x

@jumpman8282 5 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.

@rimantasri4578 5 ай бұрын

9:52 but if you're looking for x > 0 then not only 0

@williamperez-hernandez3968 5 ай бұрын

Taking the domain x>1 does not let you take the correct limit x approaching zero.

@77Chester77 5 ай бұрын

Bravo

@michelmegabacus7894 5 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

@klementhajrullaj1222 5 ай бұрын

And if, |x^3-x|/(|x^3|-|x|), or |x^3-x|/(|x^3|-x)??? ...