How to find the equation of a line tangent to a function 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!
 A: 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 $0$. 
A: 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.
A: 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.
A: 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.
