2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $|f(x_n)-f(c)| < \epsilon'$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $n\ge N$:

$$\left|f(x_n)-f(x)\right|\le(\epsilon+f'(c))\epsilon_0\le\epsilon$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.