evaluate $\lim_{x\rightarrow 0}\frac{x}{\left | x-1 \right |-\left| x+1\right|}$

Question

I just have a quick question about limits like this one: $$\lim_{x\rightarrow 0}\frac{x}{\left | x-1 \right |-\left| x+1\right|}$$

leaving it as is i get $$\lim_{x\rightarrow 0}\frac{x}{\left ( x-1 \right )-\left( x+1\right)}$$
$$\lim_{x\rightarrow 0}\frac{x}{-2}$$

$$= 0$$

which is wrong, so by saying $\left | x-1 \right |=\left | 1-x \right |$ I get $$\lim_{x\rightarrow 0}\frac{x}{\left ( 1-x \right )-\left( x+1\right)}$$

$$\lim_{x\rightarrow 0}\frac{x}{-2x}$$

$$\lim_{x\rightarrow 0}\frac{1}{-2}$$

$$=\frac{1}{-2}$$

which, I'm pretty sure, is correct. I guess my question is, if this were an exam and I had no access to a calculator or a graphing program or something, is there anyway to intuitively see that the first, obvious, answer is incorrect?

or in a situation like this do you just have to evaluate through trial and error?

Answer 1

Do not remove the absolute value signs until you are sure if the quantity inside is positive or negative.

As you are looking for the limit as $x$ approaches zero you can tell that $x-1$ will be negative and $x+1$ will be positive. So you then know that $|x-1|=1-x$ and that $|x+1|=x+1$. These leads to your second solution.

Answer 2

another way: $\frac{x}{\sqrt{(x-1)^2}-\sqrt{(x+1)^2}} \\ =\frac{x}{\sqrt{(x-1)^2}-\sqrt{(x+1)^2}} \cdot \frac{\sqrt{(x-1)^2}+\sqrt{(x+1)^2}}{\sqrt{(x-1)^2}+\sqrt{(x+1)^2}} \\ =\frac{x [ \sqrt{(x-1)^2}+\sqrt{(x+1)^2}]}{(x-1)^2-(x+1)^2} \\ =\frac{x[ \sqrt{(x-1)^2}+\sqrt{(x+1)^2}]}{([x-1]-[x+1])([x-1]+[x+1])} \text{ note I used difference of squares formula here } \\ =\frac{ x[ \sqrt{(x-1)^2}+\sqrt{(x+1)^2}]}{(-2)(2x)} \\ =\frac{\sqrt{(x-1)^2}+\sqrt{(x+1)^2}}{-4} \text{ note this function is continuous at } x=0 $