Q:

Why does $\lim_{h \to 0} f(x+h) = f(x)$ according to Lang's Short Calculus?

On Chapter III (Derivatives), section 5 (Sums, products, & quotients) of Lang's Short Calculus, Lang writes the following excerpt:

It's unclear to me why Lang asserts $$f(x+h) = f(x) + h \frac{f(x+h)-f(x)}{h} $$
From independent research, I know this is true if the function is continuous (https://www.quora.com/Does-Lim-f-x+h-f-x-as-h-approaches-0). However, my question is how did Lang arrive at this assertion (aka why is the assertion true)? Thanks.

A:

When Lang states $f(x+h) = f(x) + h \frac{f(x+h)-f(x)}{h}$ he isn't talking about limits.  He's talking about an actual real measurable non-zero number, $h$.
And it's true: $f(x+k) = f(x) + k \frac{f(x+k)-f(x)}{k}= f(x) + k\frac{f(x+k)}k - k\frac {f(x)}{k} = f(x) + f(x+k) - f(x) = f(x+k)$
His point is that BECAUSE $f(x+k) = f(x) + k \frac{f(x+k)-f(x)}{k}$ for all $k \ne 0$ then IF $\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$ has a limit.  We know $\lim_{h \rightarrow 0} f(x + h) =$
$ \lim_{h \rightarrow 0}(f(x) + h \frac{f(x+h)-f(x)}{h})=$
$f(x) +  \lim_{h \rightarrow 0}h* \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}=$
$f(x) + 0*f'(x) = f(x)$.

