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Let:

$$g(x)=\frac{1}{1+e^{1/(x-1)}}$$

for $x\ne 1$ and $g(x)=a$ for $x=1$.

For what values of $a$ will $g(x)$ be continuous for every $x$?

Thanks in adavance!

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    See [here](http://meta.math.stackexchange.com/questions/5020/tex-latex-mathjax-basic-tutorial-and-quick-reference) for a reference to make the math in your question much more readible.2012-12-28
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    Will apply this from now on. Thanks.2012-12-28

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Hint Look at $\lim\limits_{x\rightarrow1^+}g(x)$ and $\lim\limits_{x\rightarrow1^-}g(x)$ to determine what $a$ needs to be.

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    But since limg(x) from the right is different from the left, can the function ever be continues for all x? Even if a=1, then function still isn't continues.2012-12-28
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    @pie If the two one-sided limits are not equal, what can you say about the two-sided limit, and hence the continuity of the function at that point?2012-12-28
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    The two sided limit doesn't exist - but then the questions asks "For what values of $a$ will g(x) be continues for all x?".2012-12-28
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    @pie Looking at the graph [here](http://m.wolframalpha.com/input/?i=limit+of+1%2F%281%2Bexp%281%2F%28x-1%29%29%29+as+x+approaches+1&x=0&y=0) it is immediately obvious that **no** value of $a$ will make the function continuous at $x=1$. The limit analysis that you did confirms this conclusion.2012-12-28
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    Thank you, what a terrible wording that question has.2012-12-28
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    @pie No problem.2012-12-28