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Here is the question which I am trying to solve:

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

$y=\frac{1}{1+e^{1/x}}$

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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    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.2011-12-08

2 Answers 2

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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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    Do we always take the negative of the what value limit is approaching to determine the left hand limit? Like in this case the left hand limit was 1, probably because you took -0 to plug in place if x. So, do we always take the negative of what the limit tends to?2018-06-14
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Your function doesn't pass your first criterion as $\frac 1x$ is undefined at $x=0$. There is no need to go further.