Is this allowed?

My thinking is that [math]x^4[/math] can be ignored, since it's infinitely insignificant compared to [math]x^2[/math] when x is approaching 0.

look for equivalent.

What do you mean?

NO NO FUCK NO

Take x^2 out of the root by common factor and cancel it with the denominator. It converges to 1.

that your idea is correct, and has been formalized in general fashion by the notion of "equivalent" function around a value.

Scratch that.

Yes.

never ever write again 0/0 = 1 (with whatever signs you want). 0/0 is undetermined limit.

However your idea is good. To implement it, take the predominant term and factorize it :
sqrt(x^2 (1 - x^2) ) / x = abs(x)/x * sqrt(1 - x^2 )
since x tends to 0 by negative values, abs(x) = -x, so that the expression you want to calculate the limit is

- sqrt(1 + x^2 )

and this tends to - sqrt(1) = -1

I don't think this is correct...
sqrt(x^2-x^4) = x sqrt(1 - x^2)
From there you can factor x out, leaving you with 1 - x^2. Take the limit and you get 1.

The smallness of something in calculus should not be the basis for a conclusion when you can reduce it.

* sqrt(x ^ 2) is not equal to x but to abs(x). this is high school knowledge

* there is nothing to add, limit is -1

* btw check the sign of the expression : it is negative ; how the limit could possibly be positive then ?

>sqrt(x^2-x^4) = x sqrt(1 - x^2)
That can only be done when x>0, which in this case it is not.

Its a semi circle multiplied by the sign function, so its -1.

it is. sqrt(a^2) = abs(a).

Sqrt(x^2) = +-x
Since x approaches zero from the left, it's -x

Check it yourself, your wrong..
sqrt(5^2-5^4)/5 == 5sqrt(1-5^2) =/= -sqrt(1+5^2)

Except the limit is looking at x -> 0 from the left hand side. Quit posting user. You're making yourself look like a normie who doesn't get precalculus

topkek

you apparently don't know what the absolute value of a real number is.

when x is positive, limit in 0+ is +1.
when x is negative, limit in 0- is -1.

here's a little illustration for you to understand.

go to bed now, son

When u didnt pass algebra but are in calculus. Just take out the x^2 xd

I did pass algebra but it was a long time ago.
Close enough I guess, haha.

How do you identify it as a semi circle just by looking at the equation?

>Hi I am underaged: the post

˃>8355207
>Hi my baits are super weak: the post

am i retarded? or is this actually that easy

yes

[math]\sqrt{x^2}=|x|[/math]
Since in this example x is approaching 0 from the left/negative side, |x| is negative, and the answer becomes -1.
Otherwise you are correct.

there are few numbers a and b for which sqrt(a-b) equals sqrt(a) - sqrt(b).
In precise words: you're retarded.

>otherwise you are correct
Don't fool him like that user.

I didn't fool him, I am also just retarded.

Let this thread be a testament to the faggots whining "abloobloo Veeky Forums is so pretentious and dismissive"
OP asked the most retarded question, argued against the answer and still got shit explained to him

For the record I only asked a retarded question. It was a different retard who argued against the answer.

And I suppose I would be that "retard." As I said, the answer was wrong, check yourself.

Correct. As x approaches 0, [math]x^4[/math] approaches 0 much quicker than [math]x^2[/math] so the [math]x^4[/math] term can effectively be discarded.

Why do we poison students' brains with this hand-wavey bullshit? Limits have been formalised for a good while now, maybe we should start acting like it.

here's how i did it

[math] \displaystyle \lim_{x \rightarrow 0^-} \frac{\sqrt{x^2-x^4}}{x} = \lim_{x \rightarrow 0^-} \frac{\sqrt{(x^2)(1-x^2)}}{x} = \lim_{x \rightarrow 0^-} \frac{|x|\sqrt{1-x^2}}{x} [/math]

[math] \displaystyle = \lim_{x \rightarrow 0^-} \text{sign}(x)\sqrt{1-x^2} [/math]

[math] \displaystyle \lim_{x \rightarrow 0^-} \text{sign}(x) = -1 [/math]
[math] \displaystyle \lim_{x \rightarrow 0^-} \sqrt{1-x^2} = 1 [/math]

the limit of the product is the product of the limits, so

[math] \displaystyle \lim_{x \rightarrow 0^-} \frac{\sqrt{x^2-x^4}}{x} = -1 [/math]

You dont even need to use the sign limit, as [math]x \rightarrow 0^(-)[/math] implies |x|/x = -1.

[math]x \rightarrow 0^{-}[/math]

is what I meant

I assume you are & ?
Replacing x with 5 proves nothing. Different values on x give different outcomes for the equation.

how do you guys see anything nice with the
[maths] thing?

with this :
[math] \displaystyle \lim_{x \rightarrow 0^-} \frac{\sqrt{x^2-x^4}}{x} = \lim_{x \rightarrow 0^-} \frac{\sqrt{(x^2)(1-x^2)}}{x} = \lim_{x \rightarrow 0^-} \frac{|x|\sqrt{1-x^2}}{x} [/math]

the only thing I see is fucking unreadable latex commands.

pléazzz helpeee (5th element style)

Perhaps update your browser? Switch to Chrome if you aren't using it already.

okay thanks user for answering
I'll check that out

You have to have Javascript enabled, and safari on both phones and macs don't seem to run it