How is the slope of the tangent line is lim delta f over delta x as delta x approaches 0 and delta f is fixed? I guess the line would tend to be vertical as delta x approaches to 0 in this case. So could you please clear that out for me? Thank you!

Question

phi · Answer

the slope is
$$ \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x ) - f(x)}{\Delta x }$$

using h (easier to type), think of two points at
(x, f(x) )
and (x+h,  f(x+h) )
where h is tiny (very tiny)
slope is change  in y divided by change in x  (here f(x) is the y value)
$$ m = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h}$$
if we let h approach (but never reach 0), we write:
$$f'(x)= \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$$
which is "like the slope", but for two points that are infinitesimally close to each other.
By definition, this is the derivative of f(x)

kmesbah · Answer

So basically delta f is not fixed because obviously it is equal to $$f (x _{0} + \Delta x) - f(x_{0})$$. And as delta x approaches 0 delta f also approaches 0. Thank you!