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I currently work on a problem where I consider a model which requires calculation of the term

$$\frac{1}{U'(0)}$$

with $U'(x)=x^{-0.5}$.

Therefore $U'(0)$ is undefined.

However, $\frac{1}{U'(x)}=x^{0.5}$ is defined for $x=0$ so I get

$$\frac{1}{U'(0)}=0^{0.5}=0$$

Is this conclusion allowed (even though $U'(0)$ is undefined)?

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    Can you just get away with taking the limit of $1/U'(0)$ as $x$ approaches $0$? – 2017-02-17
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    Wouldn't it be more relevant that your model uses a term $Y'(x)=x^{0.5}$ instead of $1/U'$ ? – 2017-02-17

2 Answers 2

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No. The domain of $1/U'(x)$ is everything in the domain of $U'(x)$ such that $U'(x) \ne 0$. But $0$ is not in the domain of $U'(x)$.

While it's true that $1/U'(x) = x^{0.5}$, the problem is actually getting to $x^{0.5}$. Consider: $$\frac1{U'(x)} = \frac1{\frac1{x^{0.5}}} = x^{0.5}$$

That second equality won't work if $x=0$ because the middle expression is undefined if $x=0$.

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    I did a numerical test and found that you are correct, I cannot apply it that way. – 2017-03-20
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This will depend on your model. In some cases, you know the final result is defined and continuous (for reasons based on the model). If that is true, then having a few points where the calculation is undefined will not matter.