# Simple limit returning NaN: f(x) = exp(-1/x)

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Niklas Kurz il 12 Apr 2021
Modificato: John D'Errico il 12 Apr 2021
By intuition and some opinions on Mathstack the limit of this function should be coming close to 0.
However
syms x; limit(exp(-1/x),x,0)
ans =
NaN
holds another solutions. So?
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KSSV il 12 Apr 2021
syms x
f = exp(-1/x) ;
limit(f,x,0,'left')
ans =
limit(f,x,0,'right')
ans =
0

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### Risposta accettata

John D'Errico il 12 Apr 2021
Modificato: John D'Errico il 12 Apr 2021
Think of it like this. For a limit to exist at x==0, it MUST have the same limit as x approaches 0 from above, as it does when x approaches 0 from below.
Is that true? For positive values of x, we can just try a few.
format short g
x = 10.^(-6:1)
x = 1×8
1e-06 1e-05 0.0001 0.001 0.01 0.1 1 10
exp(-1./x)
ans = 1×8
0 0 0 0 3.7201e-44 4.54e-05 0.36788 0.90484
And clearly that seems to approach 0 for small positive values of x. Intuitively, we can guess the limit is zero from above.
syms X
limit(exp(-1/X),X,0,'right')
ans =
0
As you can see, MATLAB agrees with that. But what happens when x is negative?
x = -0.1;
exp(-1/x)
ans =
22026
Now when x is a small NEGATIVE number, then -1/x is a large positive number. The limit would seem unlikely to be zero when taken from below.
limit(exp(-1/X),X,0,'left')
ans =
MATLAB agrees with that claim. So the limit is inf when viewed from the left. And as I said, if the limit differs when viewed from different directions, the limit is undefined at that point. If that is the case, then the limit is only defined when viewed from a specific direction.
So don't believe everything you read on the internet, it is not always correct. Whoever claimed that on Mathstack was simply wrong, or perhaps you read what they said incorrectly.
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### Più risposte (1)

Matt J il 12 Apr 2021
Modificato: Matt J il 12 Apr 2021
The function you have shown does not have a well-defined limit as x-->0, but this does:
syms x; limit(exp(-1/abs(x)),x,0)
ans =
0
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