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Approaching to zero, but not equal to zero, then why do the points get overlapped?

You get the derivative of $f(x)$ by getting

the limit as $h$ tends to $0$ of $\dfrac{f(x+h) - f(x)}{(x+h) - (x)}$

I understand that the value of the derivative is the slope of the graph of the function at $x$. However when $h = 0$ you have just one point and you need $2$ points for a slope, don't you?

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jjb
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    Which is why we take the limit rather than taking $h=0$. – Thomas Andrews Dec 03 '11 at 22:56
  • But in differentiating from first principles you treat any hs left over after cancelling as 0 – jjb Dec 03 '11 at 22:59
  • Related to this previous answer. It's essentially the same as Berkeley's objection (if it's two points, then it's not the tangent line; if it is one point, then you aren't defining a line). See if that answer settles your doubts. – Arturo Magidin Dec 03 '11 at 23:00
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    @Arturo: I ran into an old comment of me asking you why you don't have handy .pdf files prepared like Pete L. Clark; but lately it seems that at long last this website offers a suitable replacement! :-) – Asaf Karagila Dec 03 '11 at 23:02
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    @jaques: To be precise, "the slope of the graph" is defined to be "the slope of the tangent line to the graph". The tangent line to the graph is defined in terms of the relative error (see the previous answer I linked to above). The derivative at a point is defined to be the slope of the tangent line to the graph at the point, or equivalently, as the limit above, because we can show that this limit, whatever it is, must be the slope of the tangent line to the graph at that point. – Arturo Magidin Dec 03 '11 at 23:05

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Note that with one point you cannot just use a formula that uses the rate of change of $x$ because you would divide by zero. That is why we take the limit of $h$ going to zero.

Srivatsan
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