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A commonly-accepted definition of a tangent line is the following.

A tangent line is a straight line that touches a function at only one point.

However, there are clearly cases where a tangent at a point touches the function at another point. The example that comes to mind right now is $f(x) = x\sin(x),$ where the derivative at $x=0$ is $0$ and the line with zero slope with $y=0$ intersects the function at numerous (an infinite number of) places, including $(\pi, 0)$ for instance.

Is the above definition of a tangent line sufficient, and how so? If not, what is a better definition for a tangent line?

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closed as off-topic by Adam Hughes, Rob Arthan, David K, user91500, JMP Feb 14 '17 at 7:36

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  • $\begingroup$ A tangent line at a point $x=a$ is a line which touches the function at $x=a$ $\endgroup$ – user160738 Feb 13 '17 at 21:03
  • $\begingroup$ A slightly better definition would be to only consider the tangent line touching the curve once in a neighborhood of the point. $\endgroup$ – D_S Feb 13 '17 at 21:04
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    $\begingroup$ That is a totally insufficient (and incorrect) definition of a tangent line. It works only for a convex curve. And @user160738, what you wrote is not right at all. $\endgroup$ – Ted Shifrin Feb 13 '17 at 21:04
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    $\begingroup$ Possible duplicate of Tangent definition $\endgroup$ – David K Feb 13 '17 at 21:52
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    $\begingroup$ And here is some more evidence that people use this definition of tangent (presumably in an attempt to teach calculus?), showing why it is a bad definition. $\endgroup$ – David K Feb 13 '17 at 21:58
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Here is my answer from an earlier question (How is the derivative truly, literally the "best linear approximation" near a point?), which shows that the tangent is the best local linear approximation to the function at a point:

I'll first give a intuitive answer, then an analytic answer.

Intuitively, the tangent goes in the same direction as the function, following it as closely as possible for a line. Any other line immediately starts to diverge from the function.

Analytically:

Consider the Taylor aproximation at $x$: $f(x+h) =f(x)+hf'(x)+h^2f''(x)/2+... $.

This means that, for small $h$ $f(x+h) \approx f(x)+hf'(x)+h^2f''(x)/2 $ so that the error $E(x, h) =f(x+h)- (f(x)+hf'(x)) $ is about $ h^2f''(x)/2 $.

Now consider any other line through $(x, f(x))$ with slope $s$, with $s \ne f'(x)$. At $x+h$, its value is $f(x)+sh$, so its error, $e(x, h)$ is $e(x, h, s) =f(x+h)-(f(x)+sh) $.

Since $f(x+h)-f(x) \approx hf'(x)+h^2f''(x)/2 $,

$\begin{array}\\ e(x, h, s) &=f(x+h)-(f(x)+sh)\\ &\approx hf'(x)+h^2f''(x)/2-sh\\ &= h(f'(x)-s)+h^2f''(x)/2\\ \end{array} $

so that $\dfrac{E(x, h)}{e(x, h, s)} \approx \dfrac{h^2f''(x)/2}{h(f'(x)-s)+h^2f''(x)/2} = \dfrac{hf''(x)/2}{f'(x)-s+hf''(x)/2} $.

Since $s \ne f'(x)$, as $h \to 0$, the numerator of thie ratio of errors goes to zero, while the denominator stays bounded away from zero.

Therefore the error of the tangent goes to zero faster than the error in any other line through the point.

That is why the tangent is the best linear approximation to the curve.

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There are two common alternatives to this definition:

  1. A tangent line is a linear function that locally (such that there exists a neighbourhood) touches $f$ at one point.

  2. A tangent line of $f$ at $x$ is a linear function $g$ with $g(x) = f(x)$ and $g'(x) = f'(x)$.

The definition (1) fails if $f$ a linear function or linear in a neighbourhood, in which case a linear approximation of $f$ at $x$ touches $f$ at infinitely many points. Definition (2) is much better: it coincides with the Taylor polynomial of order $1$ of $f$.

(2) can also be expressed in this form: $g$ is a tangent line of $f$ at $x$ if $g$ is linear and $$ f(y) = g(y) + \mathrm O(y^2), \qquad (y \to x) $$ in which case the existence of $g$ implies the differentiability of $f$.

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    $\begingroup$ Is it even true that locally it must touch at only one point? If you consider $f(x)=x^2\sin\frac{1}{x}$ then $f'(0)=f(0)=0$, and so the "tangent line" would be $L(x) \equiv 0$. However this intersects the graph of $f(x)$ at in finite number of points in any neighborhood of $x=0$. Perhaps this example fails because the function isn't $C^1$? $\endgroup$ – Patch Feb 13 '17 at 21:46
  • $\begingroup$ @Patch That is why definition (1) fails for non-locally-convex functions. $\endgroup$ – Henricus V. Feb 13 '17 at 21:47
  • $\begingroup$ Ah, that's a more thorough explanation. Thanks. $\endgroup$ – Patch Feb 13 '17 at 21:48
  • $\begingroup$ I don't think this is very good. Usually, the derivative is defined as the slope of the tangent line, so defining tangents in terms of derivatives is... not so well done. $\endgroup$ – Simply Beautiful Art Feb 13 '17 at 21:49
  • $\begingroup$ @SimplyBeautifulArt These definitions are equivalent. Usually derivative is defined in terms of a limit but it can also be defined as the slope of a tangent line. The latter gives more intuition. $\endgroup$ – Henricus V. Feb 13 '17 at 21:50
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Yes, the above definition is incorrect. More accurately, I would think you'd need to use the following definition instead:

A tangent line is a straight line that touches a function once, locally, at a point.

However, this fails horribly for many functions, hence my second definition:

To me, it fits better in my mind if I think about it as two points, infinitely close, touching the function twice:

enter image description here

As a side note, my second definition transcends into a much better definition of tangent lines for polynomials degree greater than 1 (not linear). If $P(x)$ is a polynomial and $f(x)$ a line, then $f(x)$ is the tangent line of $P(x)$ at $x=a$ if $Q(x)=P(x)-f(x)$ has a root of multiplicity greater than or equal to 2 at $x=a$. (the idea of roots with multiplicities greater than or equal to 2 comes about from the concept "a line that crosses two points infinitely close")

A nice example: Take $P(x)=x^2$ and $f(x)=2x-1$. We see that

$Q(x)=P(x)-f(x)=x^2-2x+1=(x-1)^2$

It has a root of multiplicity two at $x=1$, thus, $f(x)$ is the tangent of $P(x)$ at $x=1$.

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    $\begingroup$ This definition is also insufficient. Consider the tangent of any linear function. $\endgroup$ – Henricus V. Feb 13 '17 at 21:31
  • $\begingroup$ @HenryW. That's what the second half of my answer is for. $\endgroup$ – Simply Beautiful Art Feb 13 '17 at 21:36
  • $\begingroup$ Interestingly, the second definition I provide leads to the derivative of polynomials, even those with rational powers. $\endgroup$ – Simply Beautiful Art Feb 13 '17 at 21:43
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yeah that definition is imprecise. A better definition is that a tangent line at a point $x_0$ is equal to $$y-f(x_0) = f'(x_0)(x-x_0)$$ i.e. it is the line constructed but taking a point and drawing a line going through that point which has the same slope as the function at that point.

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Let $f$ be a function, let $p = (x, f(x))$ be a point on the graph of the function.

Better definition: a tangent line to $f$ at $p$ is a line $\ell$ which passes through $p$, such that there exists a neighborhood $V$ of $p$ such that $\ell$ does not pass through any other points of the graph of the function in $V$. This allows you to avoid mistakes like the example you pointed out.

Best definition: a tangent line to $f$ at $p$ is a line $\ell$ passing through $p$ whose slope is $\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$, provided this limit exists.

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