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Background: I am trying to prove differentiability implies continuity in the following way:

Proof: If $f$ is differentiable at $a$, then $$\lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a} = f'(a) $$ exists and is finite.

Then for all $\epsilon > 0$, there exists $\delta > 0$ s.t. $|x-a|<\delta \implies |\frac{f(x)-f(a)}{x-a} - f'(a)| < \epsilon$.

Then somehow show $$\tag{1}\left|\frac{f(x)-f(a)}{x-a} - f'(a)\right| \geq |f(x)- f(a)|.$$

Then for all $\epsilon > 0$, there exists $\delta > 0$ s.t. $|x-a|<\delta \implies |f(x)- f(a)| < \epsilon$.

Thus $\lim\limits_{x \to a} f(x) = f(a)$ so $f$ is continuous at $a$.

I'm aware of the usual proof. It seems like (1) must be true somehow, but I don't know how to make use of the finiteness of $f'(a)$ using this approach.

I think I can safely conclude $|\frac{f(x)-f(a)}{x-a}| > |f(x) - f(a)|$ since $|x-a|$ is a small number.

Is this approach possible? If not I would really like to understand why. I want to grow in seeing dead-ends in proof-writing faster and understand how to reason that a given direction/strategy will not work. That's my ultimate goal with this question, but of course I do want to know about this particular proof.

Arctic Char
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m4th3m4t1cs
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1 Answers1

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You are asking if one has

$$ \left| \frac{f(x) - f(a)}{x-a} -f'(a)\right| \ge c|f(x) - f(a)|$$

for some positive $c$ and the answer is no in general: just take a linear function $f(x) = f(a)+ A(x-a)$ (with $f'(a) = A$). Then the LHS is zero while the RHS is non-zero (when $A\neq 0$).

If you use the Taylor expansion (well, something you might not have at this stage),

\begin{align} f(x)-f(a) &= f'(a) (x-a) + \frac{f''(x_a)}{2} (x-a)^2, \\ \frac{f(x)- f(a)}{x-a} - f'(a) &= \frac{f''(x_a)}{2} (x-a). \end{align}

for some $x_a$ between $a$ and $x$. So the two competing terms are $f'(a)$ and $f''(x_a)/2$. So your ineqaulity does not hold when (e.g.) $f$ is a $C^2$ function so that $f'(a) \neq 0$ and $f''(a) = 0$.

Arctic Char
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