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For the following graph:

The graph

We derive an equation to find the derivative (or the slope at an instant point, x) which is given by:

The Equation

However, we are not supposed to find the tangent for the point x + h as it approaches 0. We have to find it as it approaches x for that will give the tangent at that point of the curve. So, why does the limit of h approach 0 (lim h->0) instead of x (lim h->x)?

Thanks.

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    $\begingroup$ Take a look at this page for information on how to format math on this site. $\endgroup$ – Regret Mar 19 '15 at 10:32
  • $\begingroup$ As $h$ tends to $0$, $x+h$ tends to $x$. You could also write it $f'(x)=\lim_{y\to x}\dfrac{f(y)-f(x)}{y-x}$ $-$ it comes to the same thing. $\endgroup$ – TonyK Mar 19 '15 at 10:35
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    $\begingroup$ @Regret Thank You! :) $\endgroup$ – Always Learning Forever Mar 19 '15 at 10:35
  • $\begingroup$ Uh...why the downvote? $\endgroup$ – Always Learning Forever Mar 19 '15 at 10:36
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    $\begingroup$ The downvote was probably from someone not happy with your math formatting. $\endgroup$ – Arthur Mar 19 '15 at 10:37
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As $x+h\to x\implies (x+h)-x\to0\implies h\to0$. Does that make you understand now?

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If you simply plug $\color{blue}{h}=x$ in $f(x+\color{blue}{h})$, you get $f(x+\color{blue}{x}) = f(2x)$ which is not $f(x)$.

If you simply plug $\color{blue}{h}=0$ in $f(x+\color{blue}{h})$, you get $f(x+\color{blue}{0}) = f(x)$.

That should convince us somewhat that we should let $\color{blue}{h}\to 0$ and not $\color{blue}{h}\to x$

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