Can this function be defined in a way to make it continuous at $x=0$? We have $$f=\frac{x}{\vert x-1 \vert - \vert x +1 \vert}$$
If we want to "define" this function to be continuous at $x=0$, it's limit at $0$ must equal $f(0)$. So we should find this limit and assign it to be equal to $f(0)$, then the function is continuous at $0$. Since we are looking at the function when $x\to 0$, $x\neq 0$. Lets divide both sides by $x$.
$$f=\frac{x}{\vert x-1 \vert - \vert x +1 \vert}=\frac{1}{\frac{\vert x-1 \vert}{x}-\frac{\vert x+1\vert}{x}}=\frac{1}{\vert 1-\frac{1}{x}\vert - \vert 1+ \frac{1}{x}\vert }$$
We can use $\lim \phi(x)^{-1}=\frac{1}{\lim \phi(x)}$ here ( the limit $\neq$ 0, by hypothesis ). The inverse of the limit of $\phi(x)=\vert 1 - \frac{1}{x} \vert-\vert 1+\frac{1}{x}\vert$, when $x\to 0$. If $x<1$, we have that $$\frac{1}{x}>1\implies0>1-\frac{1}{x}\implies \Bigg\vert 1-\frac{1}{x}\Bigg\vert=-\Big(1-\frac{1}{x}\Big)$$
Now if $x>0$, we have that $$\Bigg\vert 1 - \frac{1}{x} \Bigg\vert-\Bigg\vert 1+\frac{1}{x}\Bigg\vert=-2$$
and if $x<0$, then $$\Bigg\vert 1 - \frac{1}{x} \Bigg\vert-\Bigg\vert 1+\frac{1}{x}\Bigg\vert=1-\frac{1}{x}-1-\frac{1}{x}=\frac{(-2)}{x}$$
The limit of $f$ when $x\to 0$ appears to be $\frac{-1}{2}$. Could anyone tell me what errors I made in the limit finding process?
 A: There is an error at the end of your calculations. In the case $-1 < x < 0$ we have
$$
\left\vert 1 - \frac{1}{x} \right\vert-\left\vert 1+\frac{1}{x}\right\vert=
\left(1-\frac{1}{x}\right)+\left(1+\frac{1}{x}\right)=2
$$
because the argument of the first absolute value is positive, and the argument of the second absolute value is negative.
A slightly simpler approach:
Since you are interested in the limit at $x=0$ it suffices to consider $x = (-1, 1)$, $x\ne 0$. For these arguments is $x-1 <0$ and $x+1> 0$, and therefore
$$
f(x)=\frac{x}{\vert x-1 \vert - \vert x +1 \vert} = \frac{x}{-(x-1)  - ( x +1)} = \frac{x}{-2x} =-\frac 12
$$
so that 
$$
 \lim_{x \to 0 }f(x) = -\frac 12 \, .
$$
A: You get the correct result but the derivation is incorrect (or perhaps just unclear to me). You are performing the correct calculation for the $x>0$ case, but the $x<0$ should be more precise.
The second line should be:
\begin{equation}
\frac{1}{\frac{\left|x-1 \right|}{x} - \frac{\left|x+1 \right|}{x} } =  \frac{sgn(x)}{\frac{\left|x-1 \right|}{|x|} - \frac{\left|x+1 \right|}{|x|} } = \frac{sgn(x)}{\left| 1 -\frac{1}{x} \right| - \left| 1 +\frac{1}{x} \right|},
\end{equation}
where $sgn(x)$ is the sign function, which is $1$, when $x>0$, and $-1$, when $x<0$.
When $x<0$ and close to zero $\left(1 + \frac{1}{x}\right)<0$. Hence 
\begin{equation}
\left|1+ \frac{1}{x}\right| =- \left(1+ \frac{1}{x} \right).
\end{equation}
Thus in the case of $lim_{x\rightarrow0^-} f(x)$ we have
\begin{equation}
lim_{x\rightarrow0^-} \frac{sgn(x)}{\left|1 - \frac{1}{x} \right| - \left| 1 + \frac{1}{x} \right|}=lim_{x\rightarrow0^-} \frac{-1}{\left( 1 - \frac{1}{x} \right) + \left( 1 + \frac{1}{x} \right)} = -\frac{1}{2}.
\end{equation}
So the limits on both sides match, hence the function is continuous. You have missed two minus signs. One from the $sgn(x)$ at first, and another one when computing $\left| 1+ \frac{1}{x} \right|$ so your result ends up correct but only because you got lucky in getting an even number of minus signs wrong.
As a final comment, I will say that in order to see that $\left(1+ \frac{1}{x} \right)<0$ for small negative x, we can observe that $\lim_{x\rightarrow 0^-}\frac{1}{x} \rightarrow - \infty$.
A: Another way could be:
Multiply by $\frac{|x-1|+|x+1|}{|x-1|+|x+1|}$, in this way you have at the denominator:
$$(|x-1|-|x+1|)(|x-1|+|x+1|)=(x-1)^2-(x+1)^2=-4x$$
and at the nominator:
$$x(|x-1|+|x+1|)$$
you can simplify the $x$ and it remains:
$$\frac{|x-1|+|x+1|}{-4}$$
and the limit for the numerator is $2$.
