# a hint for a question: "proof an attribute of $f$ such that it's differentiable at a neighbourhoods of $0$ and $f'$ isn't continuous at $0$.."

The question is:

Suppose that $$f$$ is differentiable at a neighbourhood of $$0$$ and $$f'$$ isn't continuous at $$0$$. Show that there is exists sequences $$x_n$$ , $$y_n$$ such that:
$$y_n \xrightarrow{n\to\infty} 0 , x_n \xrightarrow{n\to\infty} 0.$$ Moreover: $$\forall n,\ x_n \neq y_n,\ x_n \neq 0 ,\ y_n \neq 0$$.
Such that:

$$\lim_{n\to\infty}\frac{f(x_n)-f(y_n)}{x_n - y_n} \neq f'(0).$$

I barely know how to begin, it's kinda trivial to show that there is exists sequence such that $$0 \neq x_n \xrightarrow{n\to\infty} 0$$ and $$\lim_{n\to\infty} f'(x_n) \neq f'(0)$$ but I don't know how does it help me, and how to continue.

I'll be glad if someone will give me a hint on how to "attack" that question.

Check that the problem is equivalent to finding a sequences $$(x_{n})_{n}$$ and $$(h_{n})_{n}$$ converging to $$0$$ such that $$x_{n} \neq 0$$, $$h_{n} \neq 0$$, $$h_{n} \neq x_{n}$$ and $$\frac{f(x_{n} + h_{n}) - f(x_{n})}{h_{n}} \not\to 0. \tag{\ast}$$

(In all my $$\to$$, it will be understood that I'm talking as $$n \to \infty$$.)

As you have noted, there exists a nonzero sequence $$(x_{n})_{n}$$ such that $$x_{n} \to 0$$ and $$f'(x_{n}) \to f'(0)$$. (This is because $$f'$$ is not continuous at $$0$$.)

Construct a sequence $$h_{n}$$ as follows: For each $$n \in \Bbb N$$, pick $$h_{n} > 0$$ satisfying the following conditions:

1. $$|h_{n}| < \frac{1}{n}$$,
2. $$\left\lvert\frac{f(x_{n} + h_{n}) - f(x_{n})}{h_{n}} - f'(x_{n})\right\rvert < \frac{1}{n}$$,
3. $$h_{n} \neq x_{n}$$.

(The existence of such an $$h_{n}$$ follows essentially by the definition of the derivative.)

By Point 1., $$h_{n} \to 0$$. Point 3. also ensures that $$h_{n}$$ has the desired structure. We only need to check that $$(\ast)$$ holds. Note that Point 2. tells us $$\frac{f(x_{n} + h_{n}) - f(x_{n})}{h_{n}} - \frac{1}{n} < f'(x_{n}) < \frac{f(x_{n} + h_{n}) - f(x_{n})}{h_{n}} + \frac{1}{n}.$$ Now, if $$\frac{f(x_{n} + h_{n}) - f(x_{n})}{h_{n}} \to f'(0)$$ were true, then by Sandwich theorem, the above would give us $$f'(x_{n}) \to f'(0)$$, contradicting our choice of $$(x_{n})_{n}$$.

Note. I'm not writing $$\lim_{n \to \infty} \cdots \neq f'(0)$$ because that should be used when the limit does exist and does not equal $$f'(0)$$. However, in the above, I simply say that the limit either does not exist or does exist and does not equal $$f'(0)$$.