What you want to be true is not. Unfortunately there is no simple connection between one-dimensional limits and $n$-dimensional limits. Well, not in a calculational sense. They appear the same in the $\epsilon, \delta$ terminology. Just replace an $\epsilon$-neighborhood with an $\epsilon$-ball.
$\lim_{x \rightarrow x_o}f(x) = L$ iff for each $\epsilon >0$ there exists $\delta >0$ such that $x$ with $0< ||x-x_o|| < \delta$ implies $||f(x)-L|| <\epsilon$.
For $n=1$, $||a|| = \sqrt{a^2} = |a|$. For $n > 1$, $||a|| = \sqrt{a_1^2+a_2^2+ \cdots a_n^2}$.
There are examples where $f(x,y) \rightarrow L$ along all possible lines through $(x_o,y_o)$ yet $\lim_{(x,y) \rightarrow (x_o,y_o)}f(x,y)$ does not exist. Similar examples exist for families of parabolic or cubic paths. It's subtle.
Now, if you know all possible paths give the same limit then you can conclude the multivariate limit exists. I have an animation illustrate the failure of linear and parabolic probing at http://www.supermath.info/ZooOfMathematicalCreatures.html and around page 11 of http://www.supermath.info/MultivariateCalculus2011Chapter3.pdf I show the calculational details. I also explain a few positive results about how to prove some limits do indeed exist.