Is there a differentiable function that approaches infinity but has a bounded derivative? I am looking for a function $\,f : \mathbb{R} \rightarrow \mathbb{R}$ that is differentiable except at one point $x$ at which it approaches infinity. Furthermore the derivative of $\,f\,$ should be bounded in a neighborhood around $x$ and not approach infinity as one approaches $x$. Is there such a function, or is this requirement contradictory?
 A: No, it is not possible. WLOG, assume that the point at which the function is 'infinite' is at 1. Then within a neighborhood of 1, the function takes on arbitrarily large values. Now, suppose that at 0, the value of the function is $c$. And suppose that the bound on the derivative within 2 of 1 is $K$.
Then the largest possible value for the function at 1 is $c + (1-0)K$ - but this is contradictory, as it's supposed to be infinite.
So it is not possible for such a function to exist.
A: Suppose the point where the function is not differentiable is $0$.  If $\lim_{x \rightarrow 0}f(x)=\infty$, we can find an increasing sequence $x_m, x_{m+1}, \ldots $ so $ \lim_{n\rightarrow \infty}x_n = 0$ and $f(x_n)=n$.  Now the mean value theorem says for all $n>m$ there is a $c_n$ so
\begin{equation}
\frac{1}{x_{n+1}-x_n}=f'(c_n).
\end{equation}
This contradicts the fact that the derivative is bounded since $\lim_{n \rightarrow 0}(x_{n+1}-x_n) =0.$
A: It is not possible. Fix a real number $y <x$ in the neighbournood of $x$ where $f^{\prime}$ is bounded. Since $\lim_{t \to x} f(t) = \infty,$ we may choose a sequence of $(y_n)$ of points with $y < y_n < x$ and $\lim_{n \to \infty} f(y_n) = \infty.$ Then for each $n,$ there is (by the mean value theorem), a point $u_n \in (y,y_n)$
with $f^{\prime}(u_n) = \frac{f(y_n) - f(y)}{y_n - y} > \frac{f(y_n) - f(y)}{x-y}.$
Hence we have $f^{\prime}(u_n) \to \infty$ as $n \to \infty,$ a contradiction, since each $u_n$ is in the chosen neighbourhood for each $n.$
A: I'm not going to give a completely rigorous proof, but I'm appealing to the idea of the derivative of $f$ at $x_0$ as the slope of the line tangent to the graph of $f$ at the point $(x_0,f(x_0))$.
Up to translating right-left and up-down, we may assume that $f(0)=0$. Let $c$ such that $|f^\prime(x)|<c$ for all $x>0$ in the connected component of the domain of $f$ containing $0$. Then the graph of $f$ must be contained in the region bounded by the lines $y=\pm cx$ and as such cannot go to infinity anywhere.
