Is this argument in the proof of differentiability implying continuity valid? This is meant to be a very straightforward proof from an elementary analysis text, but I am not sure if the justification given in the proof is correct.
The statement to prove is as follows: Let $f$ be defined on an open interval $I$, and let $c \in I$. If $f$ is differentiable at $c$, then $f$ is also continuous at $c$.
The proof given is as follows:
If $f$ is differentiable at $c$, then there is some number $f'(c)$ such that $$\lim_{x\rightarrow c} \frac{f(x)-f(c)}{x-c} = f'(c)$$
It follows that $$ \begin{align} \lim_{x\rightarrow c} (f(x) - f(c)) &= \lim_{x\rightarrow c}\biggl(\frac{f(x)-f(c)}{(x-c)}(x-c)\biggl) \\ & = \lim_{x\rightarrow c} \frac{f(x)-f(c)}{(x-c)} \times \lim_{x\rightarrow c}{(x-c)}\\ &= f'(c) \times 0 \\ &= 0 \end{align}$$
Hence by the sum and multiple rules* for limits,
$$\lim_{x \rightarrow c} f(x) = f(c)$$
This starred * comment refers to an earlier proved results about limits of functions which states that IF $\lim_{x \rightarrow c} f(x) = l$ and $\lim_{x \rightarrow c} g(x) = m$, then $\lim_{x \rightarrow c} (f(x)-g(x)) = l - m$
Am I correct in thinking that using this result is invalid? For example, how do we know a priori that $\lim_{x \rightarrow c} f(x)$ exists? (If it exists then we can apply the limit rule, but before proving this I don't see how we can use the converse of the limit rule, which doesn't seem to be true to me?)
My reasoning to complete the proof would be to use the sequential definition of a limit of a function: Since $\lim_{x\rightarrow c} (f(x) - f(c)) = 0$, we have that for an arbitrary sequence $\{x_n\}$ such that $x_n$ converges to $c$, the sequence $\{f(x_n) - f(c)\}$ converges to $0$ (by definition). Since $f(c)$ is a constant, we can conclude that the sequence $\{f(x_n)\}$ converges to $f(c)$. THEN, since $\{x_n\}$ was arbitrary, this is true for every sequence converging to $c$, and by the sequential definition of limits of a function, we can conclude that $\lim_{x \rightarrow c} f(x) = f(c)$
Is my previous paragraph correct or is it pedantry? I can see the argument seems very similar, but I am trying to be as precise as possible, and I can't quite see how using the limit rule actually applies in this case (at least not without additional reasoning).
 A: This is a reasonable concern, and the answer depends upon how you derived $\lim_{x\to c} f(x) = f(c)$ from the previous calculation that $\lim_{x\to c} \big( f(x)-f(c) \big) = 0$.
If you wrote $0 = \lim_{x\to c} \big( f(x)-f(c) \big) = \lim_{x\to c} f(x) - \lim_{x\to c} f(c) = \lim_{x\to c} f(x) - f(c)$ and solved for $\lim_{x\to c} f(x)$, this is technically incorrect for the reason you noted.
However, this is easily fixed: instead we write $f(x) = \big( f(x)-f(c) \big) + f(c)$. Both functions on the right-hand side have limits ($0$ and $f(c)$, respectively) as $x\to c$. Therefore, the sum rule for limits establishes the existence of $\lim_{x\to c} f(x)$ in addition to finding that its value equals $0+f(c)$.
A: Greg Martin's answer sums it up perfectly. For the sake of contribution, I'll provide another approach that uses the $(\varepsilon,\delta)$ definition of a limit.
We know that $\lim_{x\to c}(f(x)-f(c))=0$, so for all $\varepsilon >0$, there exists a $\delta >0$ such that for all real $x$,
$$0<|x-c|<\delta\Rightarrow|f(x)-f(c)-0|<\varepsilon$$
It is clear that $|f(x)-f(c)-0|=|f(x)-f(c)|$, so we can write
$$0<|x-c|<\delta\Rightarrow|f(x)-f(c)|<\varepsilon$$
This is precisely the statement that $\lim_{x\to c}f(x)=f(c)$.
