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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?

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3 Answers 3

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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 $$ \lim_{x \to 0 }f(x) = -\frac 12 \, . $$

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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 $\lim_{x\rightarrow 0^-}\frac{1}{x} \rightarrow - \infty$.

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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 $2$.

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