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Answer by AsdrubalBeltran for Proving that there is only one limit of a...
Hint: supposed that $L_1\neq L_2$ take $\displaystyle\epsilon=\frac{|L_1-L_2|}{2}$ and have a contradiction.
View ArticleAnswer by Siminore for Proving that there is only one limit of a function...
No: "We want to find $\lvert L_1 -L_2 \rvert =0$ or $\lvert L_1-L_2 \rvert< \epsilon$ for some small value of $\epsilon>0$" is wrong. The condition $\lvert L_1-L_2 \rvert< \epsilon$ must hold...
View ArticleProving that there is only one limit of a function using epsilon delta
Suppose $f:D \rightarrow R$ with $x_0$ as an accumulation point of D. Assume $L_1$ and $L_2$ are limits of $f$ at $x_0$. Prove $L_1=L_2$If $x_0$ is an accumulation point of D, then $\exists y \ne x_0 $...
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