Why does $f(x + \Delta x)=f(x)$ I'm just starting calculus, and was wondering if it is correct that $$f(x + \Delta x)=f(x)$$ and if so why? I'm looking for an intuitive understanding of how the variables relate.
 A: Note that in general $f(x+\Delta x) \neq f(x)$. Though the quantity $f(x+\Delta x)$ can be estimated in terms of $f(x)$.
For differentiable functions, the quantity $f(x + \Delta x)$ can be estimated by using a linearization of the function at $x$. This is a little bit better than saying that $f(x + \Delta x)$ is about the same as $f(x)$, which is true for small $x$. Using the linearization to estimate a function at $f(x+\Delta x)$ provides a slightly better accuracy.
This is the difference between saying $f(x+\Delta x) \approx f(x)$ (constant estimation) vs $f(x+\Delta x) \approx f(x) + f'(x) \Delta x$ (linear estimation).

For instance we can estimate the square root of $17$ by using the linearization of $f(x) = \sqrt{x}$ at $x=16$. Note that $f(16)=4$ and $f'(x) = \frac{1}{2\sqrt{x}}$ so that $f'(16) = \frac{1}{8}$. Thus $f(17) = f(16 + 1) \approx f(16) + \frac18 \cdot 1 = 4.125$.

When a function is differentiable we have, $$f(x+\Delta x) = f(x) + \xi \cdot \Delta x + \text{junk}$$ for some $\xi$ and where the junk term becomes very small (read: negligible) when $\Delta x$ is small. In fact if we write $\text{junk} = o(\Delta x)$ we have $o(\Delta x)/\Delta x \to 0$ as $\Delta x \to 0$.
In fact, if we write $$f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{(f(x) + \xi \Delta x + \text{junk}) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \left( \xi + \frac{o(\Delta x)}{\Delta x}\right) = \xi.$$
Thus $f'(x) = \xi$ and $f(x+\Delta x) = f(x) + f'(x) \Delta x + o(\Delta x).$

Example:
Consider $f(x) = x^2 + 2x + 1$. Then $f(x + \Delta x) = (x+\Delta x)^2 + 2(x + \Delta x) + 1$.
Using the binomial theorem (or FOIL) we have $$f(x + \Delta x) = x^2 + 2\Delta x \cdot x + \Delta x^2 + 2x + 2\Delta x + 1.$$
Collecting terms we find:
$$f(x + \Delta x) = (x^2 + 2x + 1) + (2 x + 2)\Delta x + \Delta x^2.$$
Recall that $f'(x) = 2x + 2$. Thus we have
$$f(x + \Delta x) = f(x) + f'(x) \Delta x + \Delta x^2.$$
Now notice that our junk term is $o(\Delta x) = \Delta x^2$, and $o(\Delta x)/\Delta x = \Delta x \to 0$ as $\Delta x \to 0$.
A: This is not true. I think you meant to ask if $$\lim_{\Delta x \to 0} f(x+\Delta x) = f(x).$$
The equation above is the definition of continuity at $x$. This means that if the above equation is true for some value of $x$, then the function is continuous at $x$.
For example, $f(x) = 3x+2$. This is a line (which is continuous). Notice $\lim_{\Delta x\to 0} f(x+\Delta x) = \lim_{\Delta x \to 0} 3x+3\Delta x + 2 = 3x +2$.
However, for the function $$f(x) = \begin{cases}1 & \text{ if } x \geq 0\\0&\text{ if } x< 0\end{cases},$$ it is not true that $\lim_{\Delta x\to 0} f(0+\Delta x) = f(0)$ (because the limit from the left is 0, not 1).
