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Question Number 116939 by joki last updated on 08/Oct/20

$$profifthelimitf\left(x\right)=Land$$$$limitf\left(x\right)=M,thenL=M$$

Answered by MAB last updated on 08/Oct/20

$$\underset{\mathbf{x}\to x_0}{\mathbf{lim}}f\left(x\right)=L=M$$$$\Rightarrow [(\mathrm{\forall }\epsilon >0)(\mathrm{\exists }\alpha >0)(\mathrm{\forall }x\in D_f):\mid x-x_0\mid <\alpha \Rightarrow \mid f\left(x\right)-L\mid <\epsilon ]$$$$and[(\mathrm{\forall }{\epsilon }^{\prime }>0)(\mathrm{\exists }{\alpha }^{\prime }>0)(\mathrm{\forall }x\in D_f):\mid x-x_0\mid <{\alpha }^{\prime }\Rightarrow \mid f\left(x\right)-M\mid <{\epsilon }^{\prime }]$$$$wehave\mid M-L\mid ⩽\mid f\left(x\right)-L\mid +f\left(x\right)-M\mid triangleinequality$$$$hence(\mathrm{\forall }ϵ>0)(\mathrm{\exists }\delta =min(\alpha ,{\alpha }^{\prime })>0):\mid x-x_0\mid <\delta \Rightarrow \mid M-L\mid <∍$$$$finallyM=L$$

Answered by MAB last updated on 08/Oct/20

$$\underset{\mathbf{x}\to x_0}{\mathbf{lim}}f\left(x\right)=L=M$$$$\Rightarrow [(\mathrm{\forall }\epsilon >0)(\mathrm{\exists }\alpha >0)(\mathrm{\forall }x\in D_f):\mid x-x_0\mid <\alpha \Rightarrow \mid f\left(x\right)-L\mid <\epsilon ]$$$$and[(\mathrm{\forall }{\epsilon }^{\prime }>0)(\mathrm{\exists }{\alpha }^{\prime }>0)(\mathrm{\forall }x\in D_f):\mid x-x_0\mid <{\alpha }^{\prime }\Rightarrow \mid f\left(x\right)-M\mid <{\epsilon }^{\prime }]$$$$wehave\mid M-L\mid ⩽\mid f\left(x\right)-L\mid +f\left(x\right)-M\mid triangleinequality$$$$hence(\mathrm{\forall }ϵ>0)(\mathrm{\exists }\delta =min(\alpha ,{\alpha }^{\prime })>0):\mid x-x_0\mid <\delta \Rightarrow \mid M-L\mid <∍$$$$finallyM=L$$

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