Limit theorems, prove function has a limit at every point Suppose that $f:R\to R$ is a function such that $f(x+y)=f(x)+f(y)$ for all $x,y∈R$.
Assume that $f$ has a limit at $0$, $f(1)=1$.
Prove that $f(x)=x$ for all $x \in R$
Hint: Show first that $f$ is continuous at any point $c ∈ R$. Prove then that for any rational number $r$, $f(r) = r$, and deduce the statement using the continuity of $f$.
My attempt:
To show continuity at any point $c\in R$ 
$$\lim_{x \to c}f(x)=\lim_{ h\to0}f(c+h)=\lim_{h\to0}f(h)+f(c)=f(c)$$
Suppose $\lim_{x\to0}f(x)$ exists (as given). We need to show that for $c∈R$ arbitrary, also $\lim_{x\to c}f(x)$ exists.
We have
$$f(x)=f(x−c+c)=f(x−c)+f(c)$$
Letting $x→c$ and setting $y=x−c$ we see that $y→0$, so the limit in question exists by assumption and equals
$$\lim_{x→c}f(x)=f(c)+\lim_{y→0}f(y)$$
which implies $f(x)=x$ QED
My question is how to use $f(1)=1$
Then how to prove continuity for any rational number?
 A: The posted question assumes $\lim_{x\to0}f(x)=0$, but that was not given.  What was given is that $\lim_{x\to0}f(x)$ exists.  That the limit is $0$ must be proved.
The functional equation $f(x+y)=f(x)+f(y)$ entails that $f(0)$ $=f(x+(-x))$ $=f(x)+f(-x)$; hence we must have $f(-x)=-f(x)$, i.e. this is an odd function.  If an odd function $f$ has a limit $L$ at $0$, then
$$
\lim_{x\downarrow0}f(x) =\lim_{x\to0} f(x) = \lim_{x\uparrow0} f(x),
$$
but the last one is
$$
\lim_{x\uparrow0} f(x) = \lim_{x\downarrow0} f(-x) = -\lim_{x\downarrow0} f(x) = -L.
$$
So $L=-L$, and therefore $L=0$.
As to using $f(1)=1$, notice that the function $g(x)=3x$ also satisfies the equation $g(x+y)=g(x)+g(y)$ (and similarly if we had used any other number besides $3$), so certainly that condition must be used.
What the two equalities $f(x+y)=f(x)+f(y)$ and $f(1)=1$ imply is that $f(x)=x$ whenever $x$ is rational.  Let's see that first for positive rational numbers.
\begin{align}
1 =f(1) & = f\left(\frac15+\frac15+\frac15+\frac15+\frac15\right) \\[10pt]
& = f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) = 5f\left(\frac 15 \right)
\end{align}
so $f\left(\dfrac15\right)= \dfrac15$.  And
$$
f\left(\frac35\right)=f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) = \frac15+\frac15+\frac15 = \frac 3 5
$$
and similarly for all other positive rational numbers.
If $x$ is irrational, look at a sequence of rational numbers approaching $x$ and then use continuity.
A: We see that $f$ is an automorphism on the group of real numbers with addition. 
It will suffice to show that only the identity function will be the only continuous automorphism that leaves $f(1) = 1$. If we didn't assume $f(1)=1$, we would have a lot more potential automorphisms.
A: *

*You only proved that $f$ is continuous at every point yet, that is
$$\lim_{x\to c}f(x)=f(c)$$
at every $c\in\Bbb R$. (So, your first line was enough here.)

*First prove that $f(1/n)=1/n$ for all $n\in\Bbb N$. Here you will have to use $f(1)=1$.

*Then it follows that $f(r)=r$ for all $r\in\Bbb Q$.

*Finally use continuity to prove $f(x)=x$ for all $x\in\Bbb R$.

