Real Analysis: The continuity of $\frac {1}{1-x}$ on the interval $[0,1).$

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Need help to prove the continuity of the function $f(x)=\frac {1} {1-x}$ on $[0,1).$ Using the epsilon-delta definition, I came to $\frac {|x-x_0|} {|1-x||1-x_0|}$ and I don't know how to proceed further. Can someone please help me with this proof?

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Let $0<\epsilon <1$. If $|x-x_0| <\epsilon \frac {(1-x_0)^{2}} 2$ then $\frac {|x-x_0| }{|1-x_0| |1-x|} <\epsilon$. To see this note that $1-x_0=(1-x)+(x-x_0)$ so $|1-x_0| \leq |1-x|+|x-x_0|$. Hence $|1-x| > 1-x_0-|x-x_0| >\frac {1-x_0} 2$. Can you continue from here?