If $f'(x)=0$, is then $f(x+dx)=f(x)$? I am always struggling with infinitesimals, and not sure I'm getting this right. The title basically states the simplest version of my question: If a function has zero slope at some point, is it correct to say that $f(x)=f(x+dx)$ for an infinitesimally small $dx$?
-- If yes, can anyone intuitively explain why I can drop the other terms of the Taylor expansion here?

Edit: It has been pointed out that "infinitesimal" really isn't a well-defined concept in the way that I've used it here, and the answer by Seth explains why. I've clarified what I'm after (without the use of "infinitesimals" in a separate question to avoid mixing too many different problems under one header.
Using the hyperreal number system defined by Robinson, which is a common context in which infinitesimals can be rigorously defined, a function $f$ is differentiable at $x$ if there is a real number $f'(x)$ such that $$f'(x)\approx\frac{f(x+\epsilon)-f(x)}{\epsilon}$$ for all infinitesimals $\epsilon$. Here, $a\approx b$ means that $a$ and $b$ differ by an infinitesimal. 

Thus, if $f$ is differentiable at $x$, this implies that $$\frac{f(x +dx)-f(x)}{dx}=f'(x)+\eta$$ for some infinitesimal $\eta$, which implies that $$f(x+dx)-f(x)=dx\,(f'(x)+\eta).$$ Since $dx$ is infinitesimal, the right side is infinitesimal, and therefore, in our alternative notation, $f(x+dx)\approx f(x)$.  

Interestingly, this is the definition of continuity: $f$ is continuous at $x$ if $f(x+\epsilon)\approx f(x)$ for all infinitesimals $\epsilon$. Thus the above argument shows that differentiability implies continuity.

The hypothesis $f'(x)=0$ was not used above.  This stronger hypothesis means that $$\frac{f(x+dx)-f(x)}{dx}\approx 0,$$ which means intuitively that $f(x+dx)-f(x)$ is so small that even after dividing it by the infinitesimal $dx$, it is still infinitesimal.
Suppose $f$ continuous differentiable at $x$, as it stated by the $f'(x)=0$. In the case or Riemann integral. 

$f'(x)=0\Rightarrow \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=0
\Rightarrow \frac{f(x+dx)-f(x)}{dx}=0\Rightarrow f(x+dx)-f(x)=0$.

Also $dx$ is not a number in the sence of real(in Riemann or differential geometry), it's more like an operator represtation. 
In a sense that "it never goes to anywhere". Suppose $m,n\in\mathbb{N}$, $m+dx"="m$, $m+n*dx"="m$ yet $0&lt; m*n*dx&lt;&lt;1$. ($"="$ means "sort of equal" in real, but notice this is very improper.)

For your question, answering in the case of real analysis, I think $f'(x)$ exists, which is differentiable at $x$, already implied that $f$ is continuous at $x$.

Therefore, for anly $\epsilon&gt;0$, there exists $\delta&gt;0$ such that whenever $|x_{new}-x|&lt;\delta$, $|f(x_{new})-f(x)|&lt;\epsilon$. Thus $f("x+dx")=f(x)$.