Let's see how smart Jow Forums really is

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wrong!!! it's 3

I dont understand the ,
Whats the "2, " and "3, " doing?

The function is based on x having imaginary parts to it -- x^2 can only be negative if x has an imaginary component, as x^2 on the reals is always non-negative.

As such, the limit of x approaching 0 is ambiguous -- it's not clear what sort of imaginary part might be included with x as it approaches 0.

based

For real numbers the function is not continuous for x=0 so the limes doesn't exist. I don't remember much about imaginary numbers, not enough to solve the problem anyway.

only a nigger would care to entertain such retardation. i'll stick to useful matters that people care about

2. It’s not 3 because x^2 can never be negative (assuming real numbers). We therefore can assume that since x^2 >= 0, and we don’t care about f(0), the limit approaching from the right and the left will both be 2, meaning that the limit is 2.

do your own homework

brainlet

For real x:
x^2 0 for all x =/= 0
Therefore, an equivalency is:
If x = 0, f(x) = 3
If x =/= 0, f(x) = 2

Trivially, as x approaches 0+ or 0-, x =/= 0, so f(x) approaches 2.

Ans. f(x) -> 2 as x -> 0

Jamal, don't expect us to do your homework for you.

Attached: looneycoon.jpg (300x227, 14K)

I think it's an if (never saw that notation in math class.
So if(x2 > 0) f(x) = 2
else f(x) = 3

back to your cuckshed, cracker

You wrote c1 twice.

The function needn't be continuous for a limit to exist. In fact, sin(x) -> undefined as x -> inf

2, since the limit at x=0 isn't dependant on f(0).
Just draw a graph, or imagine how it would look.

A function needs to be continous at the point where you take the limit.

except they are because x^2 for x=/=0 is always >0. lmao you faggots failed at solving jamal's homework

Math notation is the most retarded shit in existence, they should just replace it with pseudocode.

How is that true if the way you check for continuity is with a limit?

This

Why do you spastics insist on attempting to answer ambiguous questions.

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There is no more minimalistic way to express what's in that picture with pseudocode. It's overly obvious and symbolic.

the notation is fine but OP is shit, what does "," mean? I know it's a "for", but for fuck sakes just write it, or an if or a fucking arrow, not a fucking dot

Continuous functions don't always converge, function can be noncontinuous in point x1 if you want lim in x2 to exist, but if function is not continuous in x1 then lim in x1 does not exist.

>if function is not continuous in x1 then lim in x1 does not exist.
1/x -> inf as x -> 0+
1/x -> -inf as x -> 0-
There, limits clearly exist of a non-continuous function.

Fuck for some reason I was thinking about derivatives. Limes here is trivial, equals f(0) which is given in the problem almost directly.

>solving his homework
come on retards

Yeah I was thinking maybe you meant derivatives

hey let me have my 2 minutes of validation huh? i spent 80k on applied science major

Yeah see below I just wrote a struct integrity exam and can't think straight. Sorry for confusion.

Yeah my bad.

Actually I didn't write it clearly, didn't make sense.

If x is real, the limit is 2 because you can only approach x from the right (i.e. approaching from the left is the same as approaching from the right).
Of course f(0)=3 but that doesn't affect the limit, it only means f is not continuous at 0.

If you wanna read to much into it and take x complex (x=a+ib) the limit doesn't exist. Because you can find 2 "curves", c1=imaginary axis, c2=real axis, i.e. x-axis. Approaching 0=(0,0) through c1 (where x^2≤0) will give you 3, approaching (0,0) through c2 (where x^2>0) will give you 2. So it doesn't exist.

>For real numbers the function is not continuous for x=0 so the limes doesn't exist.
This is not necessary. A non continuous function can have limits. Maybe if it's discontinuous everywhere you could bhave an argument, something like
f(x)=
1, if x rational
2, if x irrational
This is discontinuous everywhere, because rationals and irrationals are dense in the reals (you can find a rational, and an irrational, arbitrarily close to any real).

>As such, the limit of x approaching 0 is ambiguous -- it's not clear what sort of imaginary part might be included with x as it approaches 0.
The point is for the limit to exist, it will have to be the same value for all possible combinations, i.e. for any curve that you take (passing through (0,0)) through which you approach (0,0). If you can find 2 different curves that give different values, by definition the limit doesn't exist.

>spending money on education
lmao, that's what people get for being retarded

cheers mate

This function is undefined for x

How else are you going to learn maths? Reading bland books?

>no solution to sqrt(-1)
nice middle school maths bro

The limit doesn't exist, approaching from the left the limit is 3, approaching from the right the limit is 2, but at x = 0, f(x) = 3.

>he didn't get into college for free
Come on.

Limits at a point are only defined to exist if both left and right limits are equal.

>This function is undefined for x

>This function is undefined for x

not x < 0, x2 < 0

makes sense
doubt it's about complex numbers though, it would've been made clear
more likely, just trying to teach freshers to think logically about conditions and that continuity and limits are not the same.

That is false. Look at the curve y=x^2. It's symmetric around the y axis.

>hurr durr there is order in the complex numbers with imaginary part
fucktard, that's clearly a function on reals

no such thing as free college

This.

If the function is defined as 3 for all x < 0 then its also defined for all x^2 < 0.

If you square both sides of x < 0 you get x^2 < 0.

>t. amerimutt

have fun being on food stamps

x^2

Isn't it safe to assume that x is real here?

the part about reals makes sense
But the part about curves flew over my head.

He is still wrong, he said it's 3 for x < 0, which is wrong. It's 2 for all real number except exactly 0, because negative numbers squared are positive.

>If the function is defined as 3 for all x < 0 then its also defined for all x^2 < 0.
Noooooo it's the other way around, the (-inf, 0) U (0, inf) interval is a superset of (-inf, 0).
E.g.
f(x) := 3, x < 0
is undefined for x > 0

Not in America, maybe. Here where I live you can join a university for free if you got the brains

>It's 2 for all real number except exactly 0, because negative numbers squared are positive.

The function is 3 for all x

What the ever-loving shit are you guys talking about?

The answer is 2.

Well free for me, wagies finance my education. Anyway I'd rather the taxes (lower here than in America) went towards this and 'free' health-care than be wasted on other shit like overgrown army or pathetically high corruption.

He's not wrong, those are equivalent transformations for the reals.

The equasion is defined for all x, its a piecewise equasion that defines all x 0, the entire "real line" is defined by the equasion and you don't need to understand what a complex number is to answer it.

Bait or blind, it's x^2

See

Don't worry too much about it. It's the definition of limits for multivariable functions.
If you have a single, say real, variable, all the numbers sit tightly on a line, the real line. So when you say "x approaches whatever" you mean x "travelling" on this line of real numbers in either direction.

In 2 variables, the numbers sit on the whole plane. So when you says "(x,y) approaches whatever", you mean (x,y) travelling towards whatever following *any* possible path on the plane, i.e. following any curve.

So the definition of the limit, instead of saying "approaches from left and right" - which are the only possible options for a single real variable - you say "approaches from *any* curve". I.e. to exist it has to give the same value for *any* curve that you choose to use to approach the required point. And equivalently, if you can find even 2 curves approaching through which gives a different result, the limit doesn't exist by definition.

There are *no* complex numbers in this problem since the imaginaries don't form a strict order, which

That's everyone in this shitty thread fucked then.

>squaring both sides of an inequation

A negative number squared is positive. So you were baiting after all.

only one with a brain here. Jow Forums once again proves it is full of /v/ermin, macfags, and NEET thinktards who think they are better than pajeets because they can write fizzbuzz in C

Please tell me you are just a bunch of high schoolers or first year calculus students, no way Jow Forums is this retarded

Most are probably baiting.

I'm 1st year college and I know it's 2, I don't know if saying 3 is bait, I would not be surprised if they are serious

>since the imaginaries don't form a strict order, which

this holy fuck tell me it's a circus act I'm not getting rn. chinese 8th graders solve this shit

>no way Jow Forums is this retarded
What odds will you give me that it is ?

Nice, thanks.

I wouldn't be so quick to agree since the squares of imaginaries are defined over a spiral in C.

>x^2
Tricky bastard, it's 2.

When they you get handed f(0) in the definition, why isn't lim f(x)x->0 just f(0)?

Because x approaches 0 in value but it is always different from 0.

?
The squares of imaginaries are reals and you use the real order.

Oh snap I was thinking of square roots

this entire thread is a fukcing disgrace
HONESTLY how many of you arre studying mathematics?

The function f(x) doesn't have to be defined at "a" for the limit to exist as you approach "a".

Is that really the case when the function is defined for the value x is approaching? I'm searching for the answer in the definition of limits but can't find an answer since every explanation talks about approaching a value for which the function isn't defined.

Yes I know but I'm asking about the cases when the function is defined there.

ufff....
lim x>-0 is 3
lim x>+0 is 2
lim x>0 doesnt exist

Someone correct me if I'm wrong.
If you approach x=0 from the left then x^2>0 so the limit is 2
If you approach x=0 from the right then x^2>0 so the limit is 2.
The function is not continuous at x=0 (b/c f(0) does not equal the limit at x=0) but that doesn't matter to say the limit exists