I have seen many proofs using this approach:

Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} = ‎f^{\prime} ‎(x) $$

and hence

$$ \lim_{x\to x_0} f(x) - f(x_0) = lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot lim_{x\to x_0} (x-x_0) = 0$$

We have therefore shown that, using the definition of continuous, if the function is differentiable at $x_0$, it must also be continuous.

However, I was wondering if you can use this same proof using the sequential definition of differentiability that states:

If $f$ is a function and has derivative $f'(c)$ at the point $c$ in the domain of $f$ means that if ($a_n$)$_{n=1}^{\infty}$ is any sequence converging to $c$ such that $a_n$ $\not= c$is in the domain of $f$ for all $n \in \mathbb{N},$ then: $$\left[ \frac{f(x_n)-f(c)}{x_n-c}\right]_{n=1}^{\infty}$$converges to $f'(c)$

My attempt using this definition:

$\left(\frac{f(x_n)-f(c)}{x_n-c}\right)_{n=1}^{\infty}$. Let $\epsilon >0.$ Then $|\frac{f(x_n)-f(c)}{x_n-c}-$$f'(c)$$| < \epsilon$ <=> |$f(a_n)-f(c)$|<($\epsilon + |f'(c)|$)|$a_n-c$|

I thought this could be the start to a proof similar to the one above, but I am stuck after this point. I'm not sure if I have to use the delta-epsilon or sequential definition of continuity to continue with this proof, or if there is another way. Any suggestions would be appreciated.


I presume $x_n$ is the same as $a_n$.

If $\left|\frac{f(x_n)-f(c)}{x_n-c}-f'(c)\right| < \epsilon$ for all large $n$, then the fact that $\left|\frac{f(x_n)-f(c)}{x_n-c}\right|-|f'(c)| \le \left|\frac{f(x_n)-f(c)}{x_n-c}-f'(c)\right|$ implies $|f(x_n)-f(c)| \le (|f'(c)|+\epsilon)|x_n-c|$ for all large $n$. Then taking $n \to \infty$, we have $|x_n-c| \to 0$ so $|f(x_n)-f(c)| \to 0$.

If you must use $\epsilon$-$\delta$ notation, then note that for sufficiently large $n$ we have $|x_n-c| < \frac{\epsilon'}{|f'(c)|+\epsilon}$ so that $|f(x_n)-f(c)| < \epsilon'$.

  • $\begingroup$ and |$f(x_n) - f(c)$| $\leq$ (|$f'(c)|+\epsilon$)||$x_n-c$| by the triangle inequality, right? $\endgroup$
    – Mathgirl
    Apr 28 '17 at 14:39

From there, you know that if $x_n\rightarrow x$, then $\forall\epsilon_0>0,\exists N$ such that:

$$n\ge N\implies|x_n-x|<\epsilon_0$$

Thus, by taking $\epsilon_0=\frac{\epsilon}{\epsilon+f'(c)}$,you get that for $n\ge N$:



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