Prove limit using epsilon delta?

it's not.

suppose we approach 0 along the line (t,0,t)

then the limit becomes

lim t-->0 (2 t^2) / (2 t^2) = 1

but if we approach along a different line such at (t,0,0)

the limit is;

lim t-->0 (0) / (t^2) = 0

If the limit exists then the limit must be the same on every path to the limit point.

That would be an adequate dis-proof, but you have been asked to use an epsilon delta proof. How do we do that?

In single variable analysis we said

lim x-->a f(x) = L means:

For any epsilon > 0 there exists a delta > 0 such that |x-a| < delta ==> |f(x) - L| < epsilon.

To generalize to multiple variables we replace |x-a| with d(x,a) where d is our chosen distance metric. Lets use the Euclidean metric.

If the limit exists and equals 0 then

For any epsilon < 0 there exists a delta > 0 such that d(X,0) < delta ==> |f(X) | < epsilon.

let epsilon = 0.1

for any delta>0 there exists X with d(X,0) < delta such that |f(x)| = 1

QED