Question about a step in the proof of: If $f:\mathbb R \to \mathbb R$ is differentiable, then it is continuous.

Question

If $f$ is differentiable at $x_0$, $$\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0).$$

So, $\lim_{x\to x_0} f(x)-f(x_0)=\lim_{x\to x_0} f'(x_0)(x-x_0)=0$.

Thus, $\lim_{x\to x_0} f(x)=f(x_0)$.

Thus, $f$ is continuous at $x_0$.

Since $x_0$ was arbitrary, $f$ is continuous on $\mathbb R$.

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My question is: about the step going from $\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$ to $\lim_{x\to x_0} f(x)-f(x_0)=\lim_{x\to x_0} f'(x_0)(x-x_0)$.

How are we allowed to do this?

Are we multiplying through by $x-x_0$? Is this allowed? Is it creating a new limit on the right side?

Answer 1

No, its $\lim \frac{a}{b} = c$, so that $\lim a = \lim (\frac ab b) = \lim \frac{a}{b} \lim b = c\lim b.$ We used two things, namely we are allowed to divide by $b$ when computing the limit $b\to 0$, and that limits have a product rule.

Answer 2

$$\begin{align*}\lim f(x)-f(x_0)&=\lim\frac{f(x)-f(x_0)}{x-x_0}(x-x_0)\\&=\lim\frac{f(x)-f(x_0)}{x-x_0}\lim (x-x_0)\\&=f'(x_0)\cdot0\\&=0\end{align*}$$

Answer 3

Note that differentiability is really much stronger than continuity: you require the ratio to effectively have a limit.

But it suffice for the ratio to be bounded (which happen to be the case since differentiability means $f'(x_0)$ exists and is finite).

Indeed $\left|\dfrac{f(x)-f(x_0)}{x-x_0}\right|\le M\implies |f(x)-f(x_0)|\le M|x-x_0|\to 0$ and $f$ is continuous.

Answer 4

You may do like this:

Since $$\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0),$$

then $$\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0)+\alpha,$$where $\alpha$ stands for an infinitesimal under the process $x \to x_0$.

Thus, $$f(x)-f(x_0)=f'(x_0)(x-x_0)+\alpha(x-x_0).$$

Take the limits of both sides as $x \to x_0.$ We obtain $$\lim_{x \to x_0}[f(x)-f(x_0)]=f'(x_0) \cdot 0+0 \cdot 0=0,$$ which is desired.