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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 for a reference to make the math in your question much more readible. –  Daryl Dec 28 '12 at 12:21
    
Will apply this from now on. Thanks. –  pie Dec 28 '12 at 12:24

1 Answer 1

up vote 1 down vote accepted

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. –  pie Dec 28 '12 at 12:28
    
@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? –  Daryl Dec 28 '12 at 12:32
    
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?". –  pie Dec 28 '12 at 12:34
    
@pie Looking at the graph here 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. –  Daryl Dec 28 '12 at 12:38
    
Thank you, what a terrible wording that question has. –  pie Dec 28 '12 at 12:48

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