Find the limit

This limit is at a point outside the transition point. this makes it a lot easier to compute after writing each absolute value as a piecewise function.

[srisaishravan]

I just wanna say that he is a great youtuber. He makes me understand these topics. He even responded to my comment last video. Thnak you Mr.Newtonm

[CalculusIsFun]

(Abs(2x-3) - abs(x-3))/x as x approaches zero.

Break apart into two. Notice that for the first one, there WOULD have been a jump discontinuity had the denominator been 2x - 3. Since it is not you know that whatever happens occurs away from where a jump would have occurred. This means only one case of the absolute value has to be considered. In this case, the negative case needs to be considered only since for all x near zero, the expression is negative excluding the absolute value. This same thing occurs for the second limit.

So you can disregard the absolute value and simply account for the negative cases do both absolute values.

so it simplifies to the limit of (-2x + 3 + x - 3)/x = -x/x = -1

The limit is -1.

[Grecks]

Answer: The limit exists and is -1.
When x approaches 0 the arguments of both absolute value terms are negative, so we can resolve them with a negative sign. The problem then boils down to the limit of -(x/x) as x approaches 0. Every Calculus student should be able to get at the answer.

[GreenMeansGOF]

I think the limit is -1. I’ll watch the video later.

[boguslawszostak]

In the numerator, there's a difference; it's enough to multiply both the numerator and denominator by the sum, and the problem "solves itself."
xor x<>0 = (3x-6)/(|2x-3|+|x-3|)
The resulting function is, as can be seen, continuous because the denominator is positive for every x.
hus, we can substitute x=0, and we get:
lim f(x) as c aproches to 0 is equal -6/6=-1

[explanitorium]

Personally, although it's correct, I'd have specifically considered x going to zero from the negative and positive directions

[Grecks]

You seem to like the absolute value function lately, eh? 😅

Edit: Here's another simple function which I like: Complex conjugation! Maybe we can play with it a bit?!

[JavedAlam]

We can use L'Hopitals rule to find the limit
x->0 of: (d/dx sqrt(2x-3)^2 - d/dx sqrt(x-3)^2) / d/dx x
= lim x->0 of: ((2(2x-3))/(sqrt(2x-3)^2) - (x-3)/(sqrt(x-3)^2)) / 1
= (-6/3) - (-3/3)
= -2 + 1
= -1

[RyanLewis-Johnson-wqxs]

Input
lim_(x->0) (abs(2 x - 3) - abs(x - 3))/x = -1
Result
True It’s in my head.

[holyshit]

Do you finished your algebra series
If not there are subjects like
Jordan's form, Cayley-Hamilton's theorem,
Orthogonalization (dot product, inner product, rotation matrices, reflection matrices, QR decomposition, orthogonalizathion of basis with different inner products,
<p(x), q(x)> = \int_{a}^{b}p(x)q(x)w(x)dx, for w(x) = 1/sqrt(1-x^2) we will get Chebyshov polynomials and for w(x)=1 we will get Legendre polynomials, for both Chebyshov and Legendre polynomians interval [a, b] is [-1, 1])

[FreeAcademyForMath]

Why did you say at the end that you should never plug in anything if there are absolute value bars? If the denominator was x+1 instead, then the answer would be immediate.

[chelliahRaveedrarajah]

I have a question, sir....
Factorize:
8x^6+11x^3+8

[RealQinnMalloryu]

{(2x)^2 ➖ (3)^2} ➖( {x)^2 ➖ (3)^2}/(x)^2={4x^2 ➖ 9} ➖ {x^2 ➖ 9}/x^2=5x^2 ➖{ x^0+x^0 ➖ }/x^2={5x^2 ➖ x^1}/x^2=5x^1/x^2=2x.1x^1 2.x^1 (x ➖ 2x+1).

[Maths]

Sir please do a limit question which was came in

JEE Advanced 2014 shift-1 question number 57

it's a question of a limit

lim as x-->1
[{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4

You have to find the greatest value of a

It has 2 possible answers 0 and 2

But I want the reason that why should I reject 2 and accept 0

Because final answer is 0

Please help 😢

[RyanLewis-Johnson-wqxs]

Limit[(Abs[2x-3]-Abs[x-3])/x, x->0]=-1 final answer.

[hadramyahmed]

Why does the graph show a value of -1 for x=0?

[anestismoutafidis]

The limit-value is (+0). I have calculated it with the Hospital-method and the Horner-method. [0/0=+0; ±6/0={ }]