If you plug in (x, mx) and get m, that is enought to say the function is not continuous.

]]>I asked Mathematica to compute the limits for different y-s:

Table[Limit[Sin[y]/x,x->0],{y,-10,10}]

And here's what I got(this is table with limits between y=-10 and y=10. The null is when y=0):

{∞, -∞, -∞, -∞, ∞, ∞, ∞, -∞, -∞, -∞, 0, ∞, ∞, ∞, -∞, -∞, -∞, ∞, ∞, ∞, -∞}

Interesting.]]>

(this plot isn't correct at x=0 coz' we get division error)]]>

If y=x, the limit will be lim x--->0 (x / ( x^2 - 2x + 3 ) ) = 0

What can I conclude from this?

]]>f(x,y) = Siny / x

It's required to find the limit of this function as (x,y) ---> (0,y)

So if I take y=mx, lim f(x,y) = Sin(mx) / x = m since Sin(mx) / mx = 1 ( maclaurin series )

Here I think this function has no limit since m is a parameter which belongs to R, it can take any value.

Most often, the question is to find whether a function has a limit. In this case, we're using y=x^2 as well. What we tend to pove is that the function has no limit at all.

f(x,y) = yx^2 / ( x^4 - 2yx^2 + 3y^2 ) if (x,y) ≠ (0,0)

f(x,y) = 0 if (x,y) = (0,0)

Study the limit of this function as ( x,y) ---> (0,0) using y=mx and y=x^2 where m is a real parameter. Show that f doesn't have a limit at the origin O.

for y =mx, the limit of f(x,y) = 0

for y=x^2, the limit of f(x,y) = 1/2

We deduce that the limit is not unique, hence, there's no limit.

]]>Traveling in abnormal ways is a good thing, such as x^2 and mx. However, most functions have different limits when approached from the axis. And besides, normally the axis is easier.

I normally use (x, 0) and (x, x). These are sufficient for most non-continuous functions, and they are normally easy to plug in.

]]>P.S. I know we can use polar coordinates, which is easier.

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