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How it's possible to find the equation to a line tangent to a function in a point where the derivative of the function is an indeterminate form?
I'm analyzing this function: $$y = \frac{x^2}{1+\log|x|}$$ And the first derivative is: $$y\,' = \frac{x(1+2\log|x|)}{(1+\log|x|)^2}$$ I have to find the line tangent in $x = 0$.
..but I'm looking for a method applicable to all kind of real-valued functions of real variable (or at least for a wide range of this).

Thanks in advance!

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1  
There is no "indeterminate form" in your example. –  André Nicolas Jul 4 '12 at 15:58
    
@AndréNicolas, Sorry, I forget to say in which point I have to find the tangent... –  Overflowh Jul 4 '12 at 16:12
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The function isn't even defined for $x=0$, as it stands. It's easy enough to see that it has a limit of $0$ for $x\to 0$, and you're free to extend it in that way. But what you then have is a function defined by cases ($x=0$ versus $x\ne 0$), and then you can't just assume that the symbolic derivative that works inside the $x\ne 0$ case will work across the boundary to $x=0$. You'll need to go back to the definition of the derivative, or at least find a more potent shortcut than ordinary symbolic differentiation. –  Henning Makholm Jul 4 '12 at 16:23

4 Answers 4

up vote 4 down vote accepted

Our function is undefined at $0$. However, it is clear that it approaches $0$ as $x$ approaches $0$. So we have a removable discontinuity at $x=0$. Remove it! Define our function to be $0$ at $x=0$.

Let our modified function be $f(x)$. We want to calculate $f'(0)$. Go back to the definition of derivative. Since $f(0)=0$, we want $$\lim_{h\to 0} \frac{h^2}{h(1+\log|h|)}.$$ This limit is easily seen to be $0$.

Remark: Since our original function is not defined at $0$, I think the proper conclusion is that the derivative does not exist at $0$. In principle, what we found is the derivative at $0$ of a different function. However, from the point of view of the geometry, there is really no problem at $0$.

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One can get an equation of line by y-y0=m(x-x0) where m is the line's slope and (x0,y0) is a point which the line passes through. To find a line tangent to a function's graph, you can simply use the point (x0,y(x0)) and the function's derivative at x0 as the slope. If the derivative isn't defined in that point, than it means that the function isn't differentiable in that points, which means there is no tangent line.

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Consider $y=\sqrt[3]{x}$. This isn't differentiable at $x=0$, but does have tangent line $x=0$. –  Cameron Buie Jul 4 '12 at 15:58
    
Talking about differentiablity in the edge of the domain where the function is defined is problematic, (note that the function isn't defined in x<0), since one can only find a one sided limit in that edge. The (two sided) limit isn't defined, hence we can not talk about a derivative, all the more so on a tangent line. In any case, a function can't have a tangent line in a point where the function is non differentiable. The (non formal) definition of differentiablity is "if the function can be line-wise approximated in a point". –  idan Jul 4 '12 at 16:15
    
The function is defined for $x<0$ (it's the cube root, not the square root). Functions can (though need not) still have a tangent line at a point of non-differentiability, but such a line will not be of the form $y-y_0=m(x-x_0)$. Every line in the plane can be written in the form $Ax+By+C=0$, but vertical lines cannot be written in point-slope or slope-intercept form. –  Cameron Buie Jul 4 '12 at 16:27
    
Sorry, I did have a mistake with that cube root. Still, that function is not differentiable at 0 (and doesn't have a tangent line there). –  idan Jul 4 '12 at 17:29
    
It is not differentiable, there, true, but it does have $x=0$ as a tangent line. To visualize it, what is the tangent line to $y=x^3$ at $0$? It is $y=0$. Now reflect everything over the line $y=x$. –  Cameron Buie Jul 4 '12 at 17:41

If your function is differentiable in x0, then the equation for the tangent is

$$ y = a \cdot x + b $$

where

$$ \begin{cases} a = f'(x_0) \\ b = f(x_0) - f'(x_0) \cdot x_0 \end{cases} $$

so that

$$ y = f'(x_0) \cdot x + (f(x_0) - f'(x_0) \cdot x_0) $$

If f(x) isn't differentiable in x0 it may be left or right differentiable, and you have two tangents if it's both left and right differentiable.

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Here, the domain of your function is $\Bbb R\smallsetminus\{-1/e,0,1/e\}$, and one can continuously extend that by explicitly stating that $y(0)=0$. It turns out, in fact, that this is a differentiable extension, as $$\lim_{x\to 0}\cfrac{\frac{x^2}{1+\log|x|}-0}{x}=\lim_{x\to 0}\frac{x}{1+\log|x|}=0,$$ since the numerator shrinks and the denominator grows (more negative) without bound. As a side note, explicitly stating that $y'(0)=0$ gives a continuous extension of $y'(x)=\frac{x(1+2\log|x|)}{(1+\log|x|)^2}$ (as you can check), so our derivative is also continuous on $\Bbb R\smallsetminus\{-1/e,1/e\}$.

Thus, at every point $x\in\Bbb R\smallsetminus\{-1/e,1/e\}$, there is a tangent line to the function $$y=\begin{cases}0 & x=0\\\frac{x^2}{1+\log|x|} & x\in\Bbb R\smallsetminus\{-1/e,0,1/e\}.\end{cases}$$ Of course, if you don't extend the function to be defined at $0$, then there's nothing to worry about, since you've got an explicit formula for the derivative at every point of the domain.

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