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Could someone give me tips on how to resolve this issue? I don’t even know where to start.

Let $f:X\longrightarrow \mathbb{R}$ be differentiable at point $a \in X.$ If $x_n<a<y_n$, for all $n$ and $\lim x_n=\lim y_n=a$. Prove that $\lim \left(\frac{f\left(y_n\right)-f\left(x_n\right)}{\left(y_n-x_n\right)}\right)=f'\left(a\right).$ Interpret this fact geometrically.

Thanks!

Bernard
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Fernando Sousa
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    What definition of differentiability are you using? – Robert Shore Jul 25 '21 at 21:24
  • @RobertShore I'm using that: $f'\left(a\right)=\lim _{x\rightarrow a}\frac{f\left(x\right)-f\left(a\right)}{x-a}=\lim _{h\rightarrow 0}\left(\frac{f\left(a+h\right)-f\left(a\right)}{h}\right)$ – Fernando Sousa Jul 25 '21 at 21:29
  • Do you have access to Mean Value Theorem? – Matthew H. Jul 25 '21 at 21:30
  • @MatthewPilling yes, I have. $f:\left[a,b\right]\longrightarrow \mathbb{R}$ is continuous,if $f$ is derivable at $(a,b)$, exists $c \in (a,b)$ such that $f'(c)=\frac{\left[f\left(b\right)-f\left(a\right)\right]}{b-a}$ – Fernando Sousa Jul 25 '21 at 21:40
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    Mean value theorem only proves the claim under the stronger assumption of differentiability on a neighborhood of $a$, rather than differentiability at $a$ (I think this was also the reason of Robert Shore asking what definition of differentiability is being used). – jacques Jul 25 '21 at 22:13

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Use $$ \frac{f(y_n) - f(x_n)}{y_n - x_n} = \frac{y_n - a}{y_n - x_n} \frac{f(y_n)-f(a)}{y_n - a} + \frac{a - x_n}{y_n - x_n} \frac{f(a)-f(x_n)}{a - x_n} = \gamma_n \eta_n+(1-\gamma_n) \xi_n = \gamma_n (\eta_n - \xi_n) + \xi_n, $$ where $$ \gamma_n := \frac{y_n - a}{y_n - x_n} \in (0, 1) \\ \eta_n := \frac{f(y_n)-f(a)}{y_n - a} \to f'(a) \\ \xi_n := \frac{f(a)-f(x_n)}{a - x_n} \to f'(a)$$ Now you use the fact that $$\limsup_{n\to \infty} (a_n b_n) = \limsup_{n \to \infty}a_n \; \lim_{n\to \infty} b_n $$ whenever $\lim b_n$ exists, to get $$\limsup_{n\to \infty} [ \gamma_n (\eta_n - \xi_n )] = (\limsup_{n\to\infty} \gamma_n)\cdot 0 = 0 ,$$ since $\limsup_{n\to\infty} \gamma_n \leq 1 < \infty$. and analogously for the lim inf. This shows that $\gamma_n (\eta_n - \xi_n) \to 0$, and we know $\xi_n \to f'(a)$, which proves the claim.

EDIT: I'm sorry, the claim about $\limsup (a_n b_n)$ is not true in general, see discussion here: limsup of the product of two sequences, of which one converges However, in this situation it's not too difficult to adapt the proof. Clearly, if $a_n \to 0$ and $\limsup b_n < \infty, \liminf b_n > -\infty$, we can pick, for any $\epsilon > 0$ an $N \in \mathbb N$ s.t. $$\forall n > N: b_n < \limsup_k b_k + \epsilon \quad \wedge \quad b_n > \liminf_k b_k - \epsilon \quad \wedge \quad \vert a_n \vert < \epsilon$$ Therefore, $\vert a_n b_n \vert \leq \epsilon (\vert \limsup_k b_k \vert + \vert \liminf_k b_k \vert + \epsilon ) \to 0$ as $\epsilon \to 0$.

jacques
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Here's a simpler version of @jacques's argument. It's much easier to follow if we set $y_n=a+h$ and $x_n=a-k$ with $h,k>0$, with $h,k\to 0$. Write \begin{multline*}\frac{f(a+h)-f(a-k)}{h+k} - f'(a) = \\ \frac h{h+k}\left(\frac{f(a+h)-f(a)}h - f'(a)\right) + \frac k{h+k}\left(\frac{f(a)-f(a-k)}k - f'(a)\right). \end{multline*} If you now have $\delta>0$ so that whenever $0<h,k<\delta$, we have $$\left|\frac{f(a+h)-f(a)}h - f'(a)\right|<\epsilon \ \text{and}\ \left|\frac{f(a)-f(a-k)}k - f'(a)\right|<\epsilon,$$ then you will finish immediately.

Ted Shifrin
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  • Yes, I was trying to use algebra and the triangle inequality to crank out an answer but why bother now after seeing your elegant answer, where the quantities $\frac{h}{h+k}$ and $\frac{k}{k+h}$ are amenable to analysis. (+1) – CopyPasteIt Jul 25 '21 at 23:59
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We define $u_n$ and $v_n$ by writing

$\tag 1 f(x_n) = f(a) + f'(a)(x_n-a) + u_n$ and $\tag 2 f(y_n) = f(a) + f'(a)(y_n-a) + v_n$

The existence of the derivative $f'(a)$ implies that

$\quad \displaystyle \lim \frac{u_n}{x_n-a} = 0 \; \land \; \lim \frac{v_n}{y_n-a} = 0$

Since $|y_n - x_n| \gt |x_n - a|$ and $|y_n - x_n| \gt |y_n - a|$ we must also have

$\tag 3 \displaystyle \lim \frac{u_n}{y_n-x_n} = 0 \; \land \; \lim \frac{v_n}{y_n-x_n} = 0$

Subtracting $\text{(1)}$ from $\text{(2)}$,

$\tag 4 f(y_n) - f(x_n) = f'(a)(y_n-x_n) + v_n - u_n$

Dividing by $y_n-x_n$,

$\tag 5 \displaystyle \frac{f(y_n) - f(x_n)}{y_n-x_n} = f'(a) + \frac{v_n}{y_n-x_n} - \frac{u_n}{y_n-x_n}$

Applying $\text{(3)}$ to $\text{(5)}$ we arrive at the desired result,

$\tag 6 \displaystyle \lim \frac{f(y_n) - f(x_n)}{y_n-x_n} = f'(a)$

CopyPasteIt
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