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Locate the discontinuities of the function and illustrate by graphing.

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

The function $y=\frac{1}{1+e^{1 / x}}$ is discontinuous at $x=0$ because the left- and right-hand limits at $x=0$ are different.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

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This is problem number thirty three to Stuart Calculus eighth edition Section two point fire. Locate the discontinuities of the function and illustrated by graphing. Why equals one divided by the quantity one plus eat too the one hour over X. Now we take a look at the function when we determine if there's any restrictions on X I'm Such restrictions include making sure that there are no negative numbers within possible negative numbers within yeah, square root function. Uh, or the denominator cannot be zero. In this case, we have an exponential function plus one and the denominator This exponential function is never going to be negative. Or, in our case, think of the one which means that it'LL never be zero the denominator. But we do have affection within this exponential function. One of Rex. So we already know Wonder section will be that, uh, ex cannot be zero. And we expect a discontinuity car to be an X equals zero as a result of this restriction on the domain, if we bring up a graph of this function. So this is the function one over one plus one over the one ninety one bliss eat of the one Rex. We see that this function is one have very far away from the origin amore. Approximately one half. That's what the limit is as it gets closer and closer to zero from the left. Ah, this value approaches zero. Which means that this entire function for just one when you're approaching zero from the right, however thiss function approaches. Ah, this entire function approaches zero. And so we see that there's this jump jump on just continuity current attic Siegel zero for the specific function. And we have confirmed this by graphing and we have come from the discontinuity is at X equals zero.

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