Multivariable limit: (sqrt(|x|)y) / (x^2+y^2)

[demonelite123]

lim (sqrt(|x|)y) / (x^2+y^2)
(x,y) -> (0,0)

so I've substituted y = x, y = sqrt(|x|) as well as the substitutions for polar coordinates. the function seems to approach infinite which means that the limit does not exist. the problem asks to show whether the limit exists or not and then to prove it.

i am a little unsure how to prove that it doesn't exist. in the case where i substitute y = x i get lim as (x,y)->(0,0) of sqrt(x) / 2x which simplifies to 1 / 2sqrt(x). i then say that as x approaches 0 then the quantity approaches infinite so the limit does not exist.

does that constitute an actual "proof" or is it too "hand-wavy". if so how would i truly prove this statement? thanks

 

[╔(σ_σ)╝]

The limit doesn't approach infinity,.

Try keeping x constant than then let y approach zero.

Do the same thing for y.

Or change to polar-coordinates.

[URL]http://i236.photobucket.com/albums/ff286/nfforums/NF%20smilies/x31pt9.gif[/URL]

 

[sponsoredwalk]

When you set y = √x what you've written is not correct, the algebra is wrong.

The goal of the limit is to show that all limits approach the same value, to prove that
a function doesn't have the limit L all you must do is find one limit different from the
others in this case. There are more situations than just the one you've spoken about
as well

 

[╔(σ_σ)╝]

As sponsoredwalk said the limits are different if you

Try keeping x constant than then let y approach zero.

Do the same thing for y.

And if you then set y= $$\sqrt{x}$$ and take the limit.

Different limits should be the condition you need.

 

[demonelite123]

ok so when i set y = mx i get m / ((1+m^2)sqrt(x)) and that approaches infinite as x approaches 0.

when i set y = mx^2 i get (m)sqrt(|x|) / (1 + m^2 x^2) and that approaches 0 as x approaches 0.

would this prove that the limit does not exist? thank you all for your previous replies.

 

[SammyS]

ok so when i set y = mx i get m / ((1+m^2)sqrt(x)) and that approaches infinite as x approaches 0.

when i set y = mx^2 i get (m)sqrt(|x|) / (1 + m^2 x^2) and that approaches 0 as x approaches 0.

would this prove that the limit does not exist? thank you all for your previous replies.

Yes, that should do it.

If you want one more path to take, try setting x = y². Unless I screwed up my algebra, that limit is also interesting, and confirms your conclusion.

Your limit along that path (from above) becomes: lim _y→0⁺ (y²/(y⁴+y²)).

From below, it is: lim _y→0^- (-y²/(y⁴+y²)).