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Continuous function
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====A useful lemma==== Let <math>f(x)</math> be a function that is continuous at a point <math>x_0,</math> and <math>y_0</math> be a value such <math>f\left(x_0\right)\neq y_0.</math> Then <math>f(x)\neq y_0</math> throughout some neighbourhood of <math>x_0.</math><ref>{{citation|last=Brown|first=James Ward|title=Complex Variables and Applications|year=2009|publisher=McGraw Hill|edition=8th|page=54|isbn=978-0-07-305194-9}}</ref> ''Proof:'' By the definition of continuity, take <math>\varepsilon =\frac{|y_0-f(x_0)|}{2}>0</math> , then there exists <math>\delta>0</math> such that <math display="block">\left|f(x)-f(x_0)\right| < \frac{\left|y_0 - f(x_0)\right|}{2} \quad \text{ whenever } \quad |x-x_0| < \delta</math> Suppose there is a point in the neighbourhood <math>|x-x_0|<\delta</math> for which <math>f(x)=y_0;</math> then we have the contradiction <math display="block">\left|f(x_0)-y_0\right| < \frac{\left|f(x_0) - y_0\right|}{2}.</math>
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