$f$ is continuous at $a$ iff the limit is the same along every path Let $f: \mathbb{R}^n \to \mathbb{R}^m$. Show that $f$ is continuous at $x \in \mathbb{R}^n$ iff for every continuous $\rho:\mathbb{R}^s \to \mathbb{R}^n$ satisfying $\rho (0) = x$ we have 
$$\lim_{t\to 0} (f \circ \rho )(t)=f(x)$$
This seems intuitively obvious. No matter which path $\rho$ I choose the limit must be the same, or else the limit does not exist. But I am unsure how to "formalize" this in a proof. 
 A: One direction is simple enough - it's a consequence of the composition of limits. The other direction is less straight forward. Given $f$ is not continuous at $a$, we need to construct a continuous path $\gamma$ that violates the limit condition in the question.
If $f$ is not continuous at $a$, there must be some $\varepsilon > 0$ such that for all $k \in \mathbb{N}$, there exists an $x_k \in \mathbb{R}^n$ such that $\|x_k - a\| \le \frac{1}{k}$ and $\|f(x_k) - f(a)\| \ge \varepsilon$. All we need to do is construct a continuous path $\gamma$ that passes through all of these points which then converges to $a$, and then $(f \circ \gamma)(t) \not\rightarrow a.$
Define $\phi : [1, \infty) \rightarrow \mathbb{R}^n$ as follows: given $t \in [1, \infty)$, let $k = \lfloor t \rfloor$, the floor of $t$. Then define
$$\phi(t) = \lbrace t \rbrace x_{k+1} + (1 - \lbrace t \rbrace)x_k,$$
where $\lbrace t \rbrace = t - k$, the fractional part of $t$. It takes some tedious verification, but this path is continuous, and piecewise linear. It basically connects the dots of the sequence $x_k$, taking the values $x_k$ at integer points.
Next, let $\psi : [0, 1] \rightarrow \mathbb{R}^n$ be defined by,
$$\psi(t) = \left\lbrace\begin{array}{ccc} \phi\left(\frac{1}{t}\right) & : & 0 < t \le 1 \\ a & : & t = 0 \end{array}\right..$$
Note that $\psi$ is continuous at every $t > 0$, as $\psi$ is a composition of continuous functions. We need to show that $\psi(t)$ is continuous as $t \rightarrow 0^+$.
Since $x_k \rightarrow a$ by construction, for all $\varepsilon > 0$, there exists some $N$ such that
$$k > N \implies \|x_k - a\| < \varepsilon.$$
Suppose then that $0 < t < \frac{1}{N+1}$. Let $s = \frac{1}{t}$, and $k = \lfloor{s} \rfloor$. Then,
\begin{align*}
s > N + 1 &\implies k, k+1 > N \\
&\implies \|x_k - a\| < \varepsilon \text{ and } \|x_{k+1} - a\| < \varepsilon \\
&\implies \lbrace s \rbrace\|x_{k+1} - a\| + (1 - \lbrace s \rbrace)\|x_k - a\| < \varepsilon \\
&\implies \|\lbrace s \rbrace x_{k+1} + (1 - \lbrace s \rbrace)x_k - a\| < \varepsilon \\
&\implies \|\psi(t) - a\| < \varepsilon.
\end{align*}
Therefore $\psi$ is continuous at $0$.
To finish the construction off, since you specified that $\gamma$ should have $\mathbb{R}$ as its domain, we can define $\gamma(t)$ to be $\psi(t)$ when $t \in [0, 1]$, $a$ when $t < 0$ and $x_1$ when $t > 1$.
EDIT: To show that $\phi$ is continuous at a given point $a$, seperate into two cases: $a \in \mathbb{N}$ or $a \not\in \mathbb{N}$.
Suppose first that $a \not\in \mathbb{N}$. Then, $\phi(a)$ lies on the line segment between $x_k$ and $x_{k+1}$ where $k = \lfloor a \rfloor$. Locally, it's just the equation of the line between those points, which is easy to work with. It's only when we reach $x_k$ or $x_{k+1}$ that unexpected things happen.
With that in mind, we will have to limit our $\delta$. We make sure that $\delta < \lbrace a \rbrace$ and $\delta < 1 - \lbrace a \rbrace$. If we stick to such $\delta$ values, we make sure to stay within the endpoints, and we can treat the equation like a line. So, given $\varepsilon > 0$, we can then set
$$\delta = \min \left\lbrace \lbrace a \rbrace, 1 - \lbrace a \rbrace, \frac{\varepsilon}{\|x_k - x_{k+1}\|} \right\rbrace,$$
or when $x_k = x_{k+1}$, we can leave that last term off.
If $|t - a| < \delta$, then note that $k < t < k + 1$, and hence
$$\phi(t) = \lbrace t \rbrace x_{k+1} + (1 - \lbrace t \rbrace)x_k,$$
where $k$ is as above (i.e. the one defined in terms of $a$). Therefore,
\begin{align*}
\|\phi(x) - \phi(a)\| &= \|\lbrace t \rbrace x_{k+1} + (1 - \lbrace t \rbrace)x_k - \lbrace a \rbrace x_{k+1} - (1 - \lbrace a \rbrace)x_k\| \\
&= |\lbrace t \rbrace - \lbrace a \rbrace | \|x_k - x_{k+1}\| \\
&= |(t - k) - (a - k)| \|x_k - x_{k+1}\| \\
&= |t - a| \|x_k - x_{k+1}\| \\
&< \frac{\varepsilon}{\|x_k - x_{k+1}\|} \|x_k - x_{k+1}\| = \varepsilon.\end{align*}
Note that when $x_k = x_{k+1}$, $|t - k| \|x_k - x_{k+1}\| = 0 < \varepsilon$. Thus, we have continuity at $a$.
On the other hand, suppose $a \in \mathbb{N}$. To keep the notation consistent, let's set $k = a$. Then, $a$ lies on the cusp between two line segments: from $x_{k - 1}$ to $x_k$ and from $x_k$ to $x_{k + 1}$ (unless $a = 1$).
In fact, I think this is easiest if we consider left and right continuity separately. Let's first deal with continuity from the right. Again, we must limit our $\delta$, this time by forcing $\delta < 1$, so that we only deal with the line segment from $x_k$ to $x_{k+1}$. Then, if $0 < t - a < \delta \le 1$, then $\lfloor t \rfloor = k$, so
$$\phi(t) = \lbrace t \rbrace x_{k+1} + (1 - \lbrace t \rbrace)x_k.$$
In this case, we let
$$\delta = \min \left\lbrace 1, \frac{\varepsilon}{\|x_k - x_{k+1}\|} \right\rbrace,$$
or again, we leave off the second term when $x_k = x_{k+1}$. Hence, if $0 < t - a < \delta$, then
\begin{align*}
\|\phi(t) - \phi(a)\| &= \|\lbrace t \rbrace x_{k+1} + (1 - \lbrace t \rbrace)x_k - x_k\| \\
&= \lbrace t \rbrace \|x_{k+1} - x_k\| \\
&= (t - a) \|x_{k+1} - x_k\| \\
&< \frac{\varepsilon}{\|x_{k+1} - x_k\|} \|x_{k+1} - x_k\| = \varepsilon.
\end{align*}
The continuity from the left proceeds in a similar fashion.
