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Here is the question which I am trying to solve:

Determine if the following function is continuous at $x=0$:


For continuity, we know that there are three criteria:

  1. $f(a)$ is defined
  2. limit is finite
  3. $\lim\limits_{x\to a} f(x)=f(a)$

But here can we say that left and right limit are infinity? and does it mean that because $1/x$ is infinite then limit at zero is equal to values of function at point zero namely (positive infinite)?please help me to clarify solution of this problem

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No; Does $f(0)$ defined? –  Quixotic Dec 8 '11 at 21:20
Hint: how does $1/x$ behave as $x\rightarrow 0^{+}$ and as $x\rightarrow 0^{-}$? What does this imply about the behavior of $e^{1/x}$ and the given function in those regimes? –  mjqxxxx Dec 8 '11 at 21:22
I'll just say that your function is not defined at $x=0$, so asking for continuity at that point does not make sense. However, if it were the case that the right and left limits were equal, then you could redefine your function to be equal to the original function outside 0, and equal to the limit, for $x=0$ (i.e. a function "defined by cases"). Unfortunately, as Eric shows in his answer, the left and right limit are not equal. –  M Turgeon Dec 8 '11 at 21:24

2 Answers 2

up vote 8 down vote accepted

Both the left limits and right limits as $x\rightarrow 0$ exist, and they are

\begin{align*} \lim_{x\rightarrow 0^+} \frac{1}{1 + e^{1/x}} &= 0 \newline \lim_{x\rightarrow 0^-} \frac{1}{1 + e^{1/x}} &= 1 \end{align*}

which follows from the fact that $\lim_{x\rightarrow 0^+} e^{1/x} = \infty$ and $\lim_{x\rightarrow 0^-} e^{1/x} = 0$.

Because the left and right limits are different, this function is discontinuous at $0$.

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It would be more rigorous to say that the function cannot be made continuous at 0 by extending it with an appropriate value. Arguably, since as it stands it it not defined at 0, it is "not even discontinuous" there. –  Henning Makholm Dec 8 '11 at 21:23
Yeah that's very true, its not even defined at zero. Good point. –  Eric Haengel Dec 8 '11 at 21:24
@Eric Haengel:Do we really need to show the limits as the problem fails the $f(0)$ itself –  Quixotic Dec 8 '11 at 21:25
Even though you don't have to calculate these limits, I still think it's healthy to ask the question: could I define a value at $0$ at make this work? –  Dylan Moreland Dec 8 '11 at 21:29

Your function doesn't pass your first criterion as $\frac 1x$ is undefined at $x=0$. There is no need to go further.

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