<s>[INST] evaluate $\lim_{x\rightarrow 0}\frac{x}{\left | x-1 \right |-\left| x+1\right|}$ 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?
 [/INST] 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 $  
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