(Untitled)

[viccro]

I have a quick limit question:

limit (x,y -> 0,0) [cosx - 1 - x^2/2][/x^4 +y^4]

I can't figure out if there's a way to make it definable...is it possible?

Comments

[amyura]

Lemme work on it. It's in indeterminate form [0/0], so you could use l'Hôpital's rule I think.....I just forget how to apply it in 3-space.

[kanato]

It seems like if you approach the origin along the y axis (with x=0), you're always going to get limit (y->0) 0/y^4 = 0, because y doesn't appear anywhere except that one term in the denominator.

When approaching along the x axis (y=0) you get limit (x->0) -x-2 = -∞. If you had +x^2/2 instead of -x^2/2 then your limit would be 1/24, but either way, that doesn't match the limit you get from approaching along the y-axis so unless you have a specific direction to approach from I would say that the limit DNE.It seems like if you approach the origin along the y axis (with x=0), you're always going to get limit (y->0) 0/y^4 = 0, because y doesn't appear anywhere except that one term in the denominator.

When approaching along the x axis (y=0) you get limit (x->0) -x-2 = -∞. If you had +x^2/2 instead of -x^2/2 then your limit would be 1/24, but either way, that doesn't match the limit you get from approaching along the y-axis so unless you have a specific direction to approach from I would say that the limit DNE.

[imluxionverdin]

I'm probably wrong. But if you approach along the x axis, where y=o (from either side) then you are trying to get the limit as x goes to 0 of

[cosx - 1 - x^2/2]/x^4

Now the numerator is continuous (basic continuous functions and combination rules) and cos(0) - 1 - 0^2 / 2 = 0 and the denominator is zero abd continuous at x=0, so you can apply l'Hopitals rule and evaluate

-sinx - x / 4x^3 at x=0, provided this latter limit exists.

Then apply l'Hopitals rule again, i.e. -sin(0) - 0 = 0 and also the denominator is zero at x=0, both numerator and denominator continuous at zero, so differentiate again

- cos(x) - 1 / 12x^2

Now the problem is the numerator is not zero at x=0, so you can't apply l'Hoiptals rule again ... so as far as I can see, this does not have a limit and is divergent.

Now if the denominator was x^2 instead of x^4 the original function would have a limit.

So, am I doing something wrong? It seems to me there isn't a limit?