I am adding this because some are apparently having trouble understanding how to fill the details of Bananach's idea, and it cannot be explained in full in 500-character comments. It is not really an answer to the OP in my opinion, as while Bananach's method works in theory, it is entirely impractical for actual verification that given double limit converges, which appears to be what the calcstudent is after.
Still, it is a very nice theoretical result. So I hope you will forgive me for trying to give it the hearing I think it deserves. Of course I will include my modification that addresses Hurkyl's concerns.
We start with the well-known result of Peano.
Theorem: There exists a curve $\alpha : [0,1] \to [0,1]^2$ which is both continuous and surjective.
Proofs abound. For example: http://www4.ncsu.edu/~njrose/pdfFiles/HilbertCurve.pdf
A couple of quick but useful expansions:
Corollary: For any real $a,b,c,d$ with $a < b, c<d$, there is a curve $\alpha_1 : [0,1] \to [a,b]\times[c,d]$ which is continuous and surjective.
Proof: The map $f : [0,1]^2 \to [a,b]\times[c,d] : (x, y) \to (a + (b-a)x,~c + (d-c)y)$ is a continuous bijection, so $\alpha_1 = f\circ \alpha$ will do.
Corollary: For any real $a,b,c,d$ with $a < b, c<d$ and points $u, v \in [a,b]\times[c,d]$ there is a curve $\alpha_2 : [0,1] \to [a,b]\times[c,d]$ wich is continuous and surjective, with $\alpha_2(0) = u$ and $\alpha_2(1) = v$.
Proof: Since rectangles are convex,
$$\alpha_2(t) = \begin{cases} u + 3t(\alpha_1(0) - u) & 0 \le t \le \frac 13\\
\alpha_1(3t - 1) & \frac 13 \le t \le \frac 23\\
\alpha_1(1) + (3t - 2)(v -\alpha_1(1)) & \frac 23 \le t \le 1\end{cases}$$
will work. (For those who may balk at the double definitions for the splices, I am asserting that the two definitions coincide, as is easily verified. This is useful for assuring continuity.)
Lemma: Let $A = \{(x,y) \in [-1,1]^2 : |x| \ge 1/2 \text{ or } |y| \ge 1/2\}$. Then there is a curve $\beta : [0,1] \to A$ which is continuous and surjective, with $\beta(0) = (1,1)$ and $\beta(1) = (1/2, 1/2)$.
Proof: Let $$B = [-1,1] \times [1/2,1]\\C = [-1, -1/2]\times[-1, 1/2]\\D = [-1/2, 1]\times[-1,-1/2]\\E = [1/2,1]\times[-1/2,1/2]$$
Then $A = B \cup C \cup D \cup E$.
Therefore, by the previous corollary, there are continuous surjective curves $\beta_B : [0,1] \to B,~\beta_C : [0,1] \to C,~\beta_D:[0,1]\to D,~\beta_E:[0,1]\to E$ with
$$\beta_B(0) = (1,1)\\\beta_B(1) = \beta_C(0) = (-1, 1/2)\\\beta_C(1) = \beta_D(0) = (-1/2, -1)\\\beta_D(1) = \beta_E(0) = (1, -1/2)\\\beta_E(0) = (1/2, 1/2)$$
Define $$\beta(t) = \begin{cases}
\beta_B(4t) & 0 \le t \le 1/4\\
\beta_C(4t - 1) & 1/4 \le t \le 1/2\\
\beta_D(4t - 2) & 1/2 \le t \le 3/4\\
\beta_D(4t - 3) & 3/4 \le t \le 1
\end{cases}$$
Continuity and surjectivity of $\beta$ are clear.
Theorem: There is a continuous surjective curve $\gamma : [0,1] \to [-1,1]^2$ such that
- $\gamma(1) = (0,0)$ and for all $t < 1, \gamma(t) \ne 0$.
- for all $n \in \Bbb N$, $\gamma([1 - 2^{-n}, 1]) = [-2^{-n},2^{-n}]^2$
Proof: Define $\gamma(1) = 0$, and for all $k \in \Bbb N$ and $t \in [1-2^{-k},1-2^{-k-1}]$, define $$\gamma(t) = 2^{-k}\beta(2^{k+1}(t-1) + 2)$$
Since $\beta \ne (0,0)$ anywhere in its range, it follows that $\gamma(t) \ne (0,0)$ for these $t$. Since $\bigcup_{k\in\Bbb N} [1-2^{-k},1-2^{-k-1}] = [0,1)$, part 1 is satisfied, provided that $\gamma$ is well-defined.
Note that if $t = 1 - 2^{-n}$, then it lies in 2 of the intervals: either $k = n$ and $t = 1 - 2^{-k}$ is the lower endpoint, or $k = n-1$ and $t = 1- 2^{-k-1}$ is the upper endpoint. All other values of $t > 0$ lie in only one interval.
If $k = n$, then $$\gamma(t) = 2^{-n}\beta(2^{n+1}(-2^{-n}) + 2) = 2^{-n}\beta(-2+2) = 2^{-n}(1,1) = (2^{-n}, 2^{-n})$$
While letting $k = n-1$ gives
$$\gamma(t) = 2^{1-n}\beta(2^n(-2^{-n}) + 2) = 2^{1-n}\beta(-1+2) = 2^{1-n}(1/2,1/2) = (2^{-n}, 2^{-n})$$
So $\gamma(t)$ is well-defined on the overlaps, and since $\gamma$ is continuous on each interval and has the same value on the overlaps, it follows that $\gamma$ is continuous everywhere on $[0,1)$.
Now if for some $n,~ 1 - 2^{-n} \le t < 1$, then there is an $k \ge n$ with $t \in [1-2^{-k},1-2^{-k-1}]$ and $$\gamma(t) \in 2^{-k}[-1,1]^2 = [-2^{-k},2^{-k}]^2 \subseteq [-2^{-n},2^{-n}]^2$$
since $[-1,1]^2$ is the codomain of $\beta$. Since $\gamma(1) = (0,0), we have $\gamma([1 - 2^{-n},1]) \subseteq [-2^{-n},2^{-n}]^2$
Conversely, if $(0,0) \ne (x,y) \in [-2^{-n},2^{-n}], let $k = min {m\mid 2^{-m-1} \le |x| \text{ or } 2^{-m-1} \le |y|}. Then $1 > |2^kx|, ~1 > |2^ky|$ (otherwise $k-1$ would be in the set), and at least one of $|2^kx|$ and $|2^ky|$ is $\ge 1/2$. So $(2^kx, 2^ky) \in A$, the codomain of $\beta$. Hence there is a $t_0 \in [0,1]$ with $\beta(t_0) = (2^kx, 2^ky)$. Let $t = 2^{-k-1}(t_0 - 2) + 1$. Then $1-2^{-k} \le t \le 1 - 2^{-k-1}$, and $\gamma(t) = (x,y)$. Since $k \ge n, 1-2^{-n} \le 1-2^{-k}$, and so $[-2^{-n},2^{-n}]^2 \subseteq \gamma([1-2^{-n},1])$, which proves condition 2.
All that remaims is to prove $\gamma$ is continuous at $1$. Let $\epsilon > 0$, and choose $n$ large enough that $[-2^{-n},2^{-n}]^2$ lies within the ball of radius $\epsilon$ about $(0,0)$. Let $\delta = 2^{-n}$. For $t\in[0,1]$, if $|1-t| < \delta$ then $t \in [1- 2^{-n},1]$, so $\gamma(t) \in [-2^{-n},2^{-n}]^2 \subseteq B_\epsilon((0,0))$ and so is continuous at $1$.
Finally, to address the question:
Theorem: Let $f$ be a real-valued function defined on a neighborhood of $(0,0)$ in $\Bbb R^2$. Then $$\lim_{(x,y)\to(0,0)} f(x,y)$$ converges if and only if $$\lim_{t\to 1} f\circ\gamma(t)$$ converges, in which case they have the same limit.
Proof: Since $\lim_{t\to1} \gamma(t) = (0,0)$, the convergence of the first limit implies the second. It remains to show that the convergence of the second implies the first.
Suppose that $\lim_{t\to 1} f\circ\gamma(t) = L$ and let $\epsilon > 0$. Then there is a $\delta_0$ such that if $1 - \delta_0 < t \le 1$ then $|f\circ\gamma(t) - L| < \epsilon$. And there is an $n$ such that $2^{-n} < \delta_0$. Let $\delta = 2^{-n}$. If $0 < |(x,y) - (0,0)| <\delta$, then in particular, $(x,y) \in [-2^{-n},2^{-n}]$ and so there is a $t \in [1- 2^{-n}, 1]$ such that $\gamma(t) = (x,y)$. Since $(x,y) \ne (0,0), t \ne 1$. So $0<|t - 1| \le 2^{-n} < \delta_0$, so $|f(\gamma(t)) - L| < \epsilon$. I.e., $|f(x,y) - L| < \epsilon$. Therefore $$\lim_{(x,y)\to(0,0)} f(x,y) = L$$ as required.
