Suppose $f:(a,b) \backslash \{c\} \rightarrow \mathbb{R}$ is a function such that
$$\lim_{x \to \ c+}f(x) \ \ \ \ and \ \ \ \ \lim_{x \to \ c-}f(x)$$both exists and are equal to a common value $l$.
Write out carefully a complete argument that in this case $\lim_{x \to \ c} f(x)=l$.

My attempt;

If $\lim_{x \to \ c+}f(x)=l$ then by definition;
$\forall \epsilon >0 \mbox{ there exists } \delta_1>0 \mbox{ such that } \forall \ \ c<x<c+\delta_1 \ \ |f(x)-f(c)|< \epsilon$

If $\lim_{x \to \ c-}f(x)=l$ then by definition;
$\forall \epsilon >0 \mbox{ there exists } \delta_2>0 \mbox{ such that } \forall \ \ c-\delta_2<x<c \ \ |f(x)-f(c)|< \epsilon$

Now take $$\delta = min\{\delta_1 , \delta_2 \}$$
$\forall \epsilon >0 \mbox{ there exists } \delta>0 \mbox{ such that } \forall \ \ c-\delta<x<c+\delta \ \ |f(x)-f(c)|< \epsilon$
This is equivalent to;
$\lim_{x \to \ c} f(x)=l$.

Is this proof correct??

  • $\begingroup$ Looks good to me. $\endgroup$ – user88595 May 8 '14 at 8:27

Almost there: you just need to replace $f(c) $ which is not defined (as $c$ is excluded from the domain of $f$) by $l$ the defined limit from below and above.


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