There is function $f(x)=e^{(c^2 - 1)/x^2}$ when $x \neq 0$ and $f(x)=c$ when $x=0$. Why is this function continuous (the $c$ is a constant)? I know that for the function to be continuous, $\lim_{x\to c}f(x)=f(c)$ but I'm not sure where to start
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$\begingroup$ Surely you have other results on continuous functions, like "The composition of two continuous functions is continuous" and "the product of two continuous functions is contunious". $\endgroup$– ArthurDec 1, 2016 at 9:16
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$\begingroup$ @Arthur Thank you for the format edit help. I'm not following what you are saying. As x approaches infinity, the limit is 1, but how do I prove it's continuous? $\endgroup$– marbleDec 1, 2016 at 9:34
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$\begingroup$ Continuous where? At what points? Clearly it's continuous at $c$, and at any $a != 0$. But what about $a = 0$? Is it true that $lim_{x\to 0} f(x) = f(0)$? That is: does $lim_{x\to 0} f(x) = c$? $\endgroup$– BrianODec 1, 2016 at 9:41
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$\begingroup$ PS As a function defined on real numbers, not "extended reals", the behavior of $f$ as $x\to+\infty$ has no bearing on the continuity of $f$ on its domain (and vice versa), as its domain contains no such thing as $+\infty$. $\endgroup$– BrianODec 1, 2016 at 9:48
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$\begingroup$ @BrianO I think I'm getting it, but for which values of c is the function continuous? How do I approach that? $\endgroup$– marbleDec 1, 2016 at 9:51
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2 Answers
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Hint:
note that, for $c^2-1=a$ we have: $$ \lim _{x \to 0}\;e^{\frac{a}{x^2}}= \begin {cases} =0 \quad \mbox{for} \quad a<0\\ =+\infty\quad \mbox{for} \quad a>0 \end{cases} $$
answered Dec 1, 2016 at 10:06
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$f\; $is continuous at $x=0 \implies \lim_{x\to 0}f(x) $ exists $\implies -1\leq c\leq 1$.
$f$ is continuous at $\mathbb R$ if $c=0\;$ or $\;c=1$.
answered Dec 1, 2016 at 9:48
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$\begingroup$ Is this correct? With $c^2 -1<0$ and $c^2 -1>0$ it is deduced that $-1<c<1$ $\endgroup$– marbleDec 1, 2016 at 10:29
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$\begingroup$ @marble My answer is correct.$c^2-1<0\iff -1<c<1$ $\endgroup$ Dec 1, 2016 at 11:52