Uni Stuff > Mathematics
Vector Calculus
QuantumJG:
I'm starting a forum for my maths subject Vector Calculus and here is my first question.
Show that the following limit does not exist:
I tried proving it by going along the path
So
Proof:
So the limit doesn't exist since the k dependency implies that the limit value will depend on what parabolic path you take.
But I'm not sure if this is right. :uglystupid2:
TrueTears:
Hi there, this is my working:
Let's approach (0,0) from the x axis.
Now let's approach (0,0) from y axis.
Seems good so far...
Now let's approach (0,0) from the straight line y = mx
So all 3 path lead to 0, now let's try parabolas.
The rest follows from your working.
It's correct.
However I'd change your final sentence to this: where C is an arbitrary constant. So we have find a path which leads to a different limit. Thus the limit at (0,0) does not exist.
To be even more rigorous, you can try a proof if you want. That'd be an overkill, but fun :P
QuantumJG:
--- Quote from: TrueTears on July 26, 2010, 10:18:55 pm ---Hi there, this is my working:
Let's approach (0,0) from the x axis.
Now let's approach (0,0) from y axis.
Seems good so far...
Now let's approach (0,0) from the straight line y = mx
So all 3 path lead to 0, now let's try parabolas.
The rest follows from your working.
It's correct.
--- End quote ---
Thanks TT!
In the proof I probably should start with linear equations and then move to higher order polynomials. Doing limits for functions of several variables is a bit more trickier because you aren't constrained to approaching the limit from the left or right, but you can approach it in more ways than you can imagine.
TrueTears:
Yup, also if y=kx^2 doesnt work try x = ky^2 and if that also leads to a same limit, you should think that the limit might actually exist and use an e-d proof to show it does.
Ahmad:
Just as an aside in the way of counter-examples, there are limits that don't exist but for which if you approach along any polynomial path you get the same value!
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