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Let $X \subseteq \mathbb{R}^n$ be an open set, $f:X \rightarrow \mathbb{R}$ continuously differentiable. Let furhter $\{x_k\}_k \subseteq \mathbb{R}^n$ be a sequence convergent to $x \in X$, $\{y_k\}_k \subseteq \mathbb{R}^n$ be a sequence convergent to $y \in \mathbb{R}^n$ and $\{t_k\}_k \rightarrow 0+$. I wanted to ask if by using the limit definition of continuity I may write:

$$\lim_{k \rightarrow \infty} \frac{f(x_k+t_ky_k)-f(x_k)}{t_k} = \lim_{t_k \rightarrow 0+} \frac{f(x+t_ky)-f(x)}{t_k}$$

, i.e.:

\begin{align}\lim_{k \rightarrow \infty} \frac{f(x_k+t_ky_k)-f(x_k)}{t_k} &= \frac{f(\lim_{k \rightarrow \infty} x_k+\lim_{k \rightarrow \infty} t_k \lim_{k \rightarrow \infty} y_k)-f(\lim_{k \rightarrow \infty} x_k)}{\lim_{k \rightarrow \infty}t_k}\\ &= \frac{f(x+\lim_{k \rightarrow \infty}t_ky)-f(x)}{\lim_{k \rightarrow \infty} t_k} = \lim_{t_k \rightarrow 0+} \frac{f(x+t_ky)-f(x)}{t_k}\end{align}

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  • $\begingroup$ Use Taylor's theorem. $\endgroup$ – copper.hat Sep 16 at 16:34
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The answer is in the positive. I will try to make a point that the continuity assumption removes many of the obstacles.


In general, it depends of how fast $t_k\rightarrow t$ and $x_x\rightarrow x$.

There is a small neighborhood $W(x)\subset X$ such that

$$ f(x+h)= f(x) + f'(x)h + r(x;h)$$

such that $\frac{\|r(x;h)\|}{\|h\|}\xrightarrow{h\rightarrow0}0$ (we might as well define $r(x;0):=0$).

Hence

$$\frac{f(x_k+t_ky_k)-f(x_k)}{t_k} =f'(x)\,y_k +\frac{r(x;x_k+t_ky_k-x)-r(x;x_k-x)}{t_k}$$

If $\|x_k-x\|=O(|t_k|)$, then $f'(x)\,y_k\xrightarrow{k\rightarrow\infty}f'(x)\,y$ and $$\begin{align} \frac{\Big\|r(x;x_k+t_ky_k-x)-r(x;x_k-x)\Big\|}{|t_k|}&\leq \frac{\|x_k+t_ky_k-x\|}{|t_k|}\frac{\|r(x;x_k+t_ky_k-x)\|}{\|x_k+t_ky_k-x\|} +\\ & \qquad\qquad \frac{\|x_k-x\|}{|t_k|}\frac{\|r(x;x_k-x)\|}{\|x_k-x\|}\\ &\leq C\frac{\|r(x;x_k+t_ky_k-x)\|}{\|x_k+t_ky_k-x\|} + C'\frac{\|r(x;x_k-x)\|}{\|x_k-x\|}\xrightarrow{k\rightarrow\infty}0 \end{align}$$ for some constants $C$, $C'$. This means that the limit on the left hand side of the OP exists and is the same as the expected directional derivative of $f$ at $x$ along $y$.


The addition of continuity on $f'$ in the OP improves things. Using the mean value theorem (the one dimensions version suffices since we are dealing with directional derivatives) we obtain $$ \Delta_k:=\frac{f(x_k+t_ky_k)-f(x_k) - (f(x+t_ky)-f(x))}{t_k}=f'(x_k+t_k\theta_k\,y_k) -f'(x+\theta'_k\,y) $$ where $0<\theta_k,\theta_k<1$. Given $\varepsilon>0$, there is $\delta>0$ such that $y\in X$ and $\|x-y\|<\delta$ implies $\|f'(x)-f'(y)\|<\varepsilon$. Since $t_k\rightarrow0$ and $y_k$ is bounded, for all $k$ large enough

$\|x_k+t_k\theta_k\,y_k-x\|,\,\|x+t_k\theta'_k\,y_k -x\|<\delta$. This shows that $\Delta_k\xrightarrow{k\rightarrow\infty}0$. As $\frac{f(x+t_ky)-f(x)}{t_k}\xrightarrow{k\rightarrow\infty}f'(x)y$, the answer to the OP is in the positive.

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