# A tangent, a few doubts

I have a question: A tangent, I am told is a line through a specified point (a, f(a)) which touches no other point in atleast one neighbourhood of 'a'. How do I prove from here that a tangent is also a line of best approximation of the function in that neighbourhood? Thanking all in anticipation :) Edit: Another question that comes to my mind is: Why can't there be more than one such lines which suffice to be the "best approximation" of the unique function ? i got this question while reading arturo's pleasant answer.

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It is not technically true that a tangent is a line which touches a curve at a single point. This is perhaps the intuitive starting point for the modern definition of a tangent but there are some issues.

If you take for example, the absolute value function, then you have many lines "touching" the function at the origin but none of these are true "tangents". On the other hand, if you take a linear function then the tangent at any point is just the function itself. In this case, the tangent and the function intersect at infinitely many points.

The answer to your question, depending on how you look at it, may be stated as a tautology. A tangent in the modern sense of the definition is a line of best approximation simply because it is defined that way.

Given a differentiable function $f$, we define the tangent line of the curve at $x_0$ to be the line passing through the point $\left(x_0,\ f(x_0)\right)$ with slope $f'(x_0)$.

So let me rephrase the question slightly and ask: Why does the derivative provide the "slope of best approximation"? The answer to that question falls under what it means for a function to be differentiable. The definition of the derivative is given as a linear approximation. The derivative of a differentiable function at a point $x_0$ is a number $f'(x_0)$ such that $$f(x_0 + h) = f(x_0) + f'(x)h + \epsilon(h)$$ where $\epsilon(h)$ is a remainder function. The remainder function represents the error in the approximation at a distance $h$ from the site of approximation and it needs to satisfy $$\lim_{h\rightarrow 0}\frac{\epsilon(h)}{h} = 0$$ So that the error is much smaller than the distance to the point of approximation. Intuitively, this condition is what characterizes the derivative as a good approximation. In this sense, the tangent is the best linear approximation because it is the only line which will satisfy the above property. The uniqueness follows from the uniqueness of a limit; the above equation implies $$\lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h} = f'(x_0) + \lim_{h\rightarrow 0}\frac{\epsilon(h)}{h} = f'(x_0)$$ so that the derivative is uniquely defined as the above limit.

It is not the best approximation in general though. You can keep adding on successive, higher order terms by picking apart the error function to get finer and finer approximations. A second order (quadratic) approximation would look like $$f(x_0 + h) = f(x_0) + f'(x_0)h + \frac{1}{2}f''(x_0)h^2 + \epsilon_2(h)$$ where $\epsilon_2$ is an even smaller remainder. You can successively define better and better approximations which leads into the concept of Taylor polynomials and Taylor series.

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+1, excellent point about tangent definition. –  Emmad Kareem Oct 28 '12 at 6:47
You need dollar signs $around the tex code. And it must satisfy this because that's more or less the definition of the derivative. – EuYu Oct 28 '12 at 8:36 i'm still thinking about the limit.. – The cat with 9 wives Oct 28 '12 at 8:37 Well, think about it this way. For the traditional derivative to hold as a limit, when we rearrange the linear approximation, we end up with right-hand side $$f'(x_0) + \frac{\epsilon(h)}{h}$$ To require that the limit exist and to be equal to$f'(x_0)$, we necessarily need $$\lim_{h\rightarrow 0}\frac{\epsilon(h)}{h} = 0$$ – EuYu Oct 28 '12 at 8:42 Well by definition, suppose $$f'(x_0) = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$$. Therefore, at$(x_0, f(x_0))$, there is line that is tangent to$f$. At$x_0$, we can explicitly write down the equation of the line through this point. You can find this line given that the slope is$f'(x_0)$. So this line is very nice to approximate the function$f\$ (which may be a really complicated one). A line is something much nice to work with!

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The question doesn't ask why a tangent is nice to work with, it asks why a tangent a nice approximation. –  EuYu Oct 28 '12 at 6:16
what should I do now? –  Chasky Oct 28 '12 at 6:18