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

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}

• Use Taylor's theorem. – copper.hat Sep 16 at 16:34

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.