Directional derivative and a modification Let $f:X \to Y$ be a (nonlinear) function between Hilbert spaces that such that
$$f(x+th)-f(x) = tD + r(t)$$
for small $t>0$, holds for a function $r$ which satisfies $t^{-1}r(t) \to 0$ as $t \to 0$.
Suppose that $b$ is such that $t^{-1}b(t) \to 0$ as $t \to 0$. Is it possible to derive an expression for
$$f(x+th+b(t))$$ in terms of the above expression involving $D$ and $r$? I guess we may need $f$ to be differentiable in all directions, which is fine. I guess I am asking whether if the directional derivative of $f$ exists at $x$, then does also the limit
$$\lim_{t \to 0}\frac{f(x+th+b(t))-f(x)}{t}$$
exist? 
Since $b(t)$ behaves like $const\times t$, I thought maybe this is true.
 A: If $f$ is Fréchet differentiable at $x$, then of course we have
$$f(x + th + b(t)) = f(x) + Df\lvert_x(th + b(t)) + r(t) = f(x) + t\cdot Df\lvert_x(h) + o(t)$$
whenever $b\colon \mathbb{R} \to X$ satisfies $\lim\limits_{t\to 0} t^{-1}b(t) = 0$.
If $f$ is only assumed Gâteaux differentiable at $x$, even if the Gâteaux differential is a continuous linear operator, bad things can happen, and
$$\lim_{t\to 0} \frac{f(x + th + b(t)) - f(x)}{t}$$
need not exist.
For an example, consider $X = \mathbb{R}^2$ and $Y = \mathbb{R}$. Let $g \colon [0, +\infty) \to \mathbb{R}$ be given by
$$g(x) = \max \:\bigl\{0, 1 - \lvert x-2\rvert\bigr\},$$
and
$$f(r\cos \varphi, r\sin \varphi) = g\bigl(r\tan\tfrac{\varphi}{4}\bigr),$$
where $r \geqslant 0$ and $\varphi \in [0,2\pi)$.
Then for every $\varphi \in [0,2\pi)$ there is a $\varepsilon > 0$ such that $f(t\cos \varphi, t\sin \varphi) = 0$ for all $t\in (-\varepsilon, +\varepsilon)$, so $f$ is Gâteaux differentiable at $0$ with directional derivative $0$ in all directions. But $f$ isn't even continuous at $0$, and if we take $h = (1,0)$ and $b(t) = t\tan (2t)\cdot (0,-1)$, then
$$\lim_{t\to 0^+} f(th + b(t)) = 1,$$
so the difference quotient blows up.
Using a smooth bump function instead of the piecewise linear $g$ from above, and a suitable amplitude depending on the angle, we can construct an example exhibiting essentially the same behaviour that is smooth ($C^{\infty}$) on $\mathbb{R}^2\setminus \{0\}$, so requiring global Gâteaux differentiability wouldn't help. For good local behaviour, one needs Fréchet differentiability.
