# What is the meaning of $\lim_{x\to0}\frac{\ln|-x|}{\ln|x|}$? [closed]

I am now working out some critical exponent, and I encountered this result $$\lim_{x\to0}\frac{\ln|-x|}{\ln|x|}.$$ Can I write this equals to 1? Here $$x=\frac{T-T_{c}}{T_{c}}$$ and $$T_{c}$$ is the critical temerature.

## closed as off-topic by Aaron Stevens, Qmechanic♦Feb 12 at 7:33

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Yes, of course you can write it equal to 1. $$\lim\limits_{x\rightarrow 0}\frac{\ln|-x|}{\ln|x|}$$ Is the same as: $$\lim\limits_{x\rightarrow 0}\frac{\ln(x)}{\ln(x)}=\lim\limits_{x\rightarrow 0}1=1$$
Because $$x$$ is positive (as long as you're not using negative temperatures, in that case you would just reverse the sign of the argument if the quotient happens to be negative).