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Can you help me with this question?

What kind of discontinuity i have at the point $x=0$, if the function defined as bellow:

$f(x)=\left\{\begin{matrix} \frac{1}{x} &x<0 \\ x^{2}& x\geq 0 \end{matrix}\right.$

I thought that the function has a discontinuity of the second kind at the point $x=0$, because at least one of the one-sided limits is infinite:

$\lim_{x \to 0^{-} }\frac{1}{x}=- \infty $

Why is it a wrong answer?

Thanks!

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planetmath.org/Discontinuous.html says you're right. But I only know of discontinuities of first and second kind (removable and essential). Maybe your teacher is using some unusual definition. –  Stuart Dec 10 '12 at 19:37
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up vote 3 down vote accepted

this is an infinite discontinuity (essential discontinuity)

Since $L_- = - \infty \ne 0 = L_+$ So at least one of the limits are undefined ($\infty$)

note: For me, this is the third type, the first 2 being a removable discontinuity (by continuous extension, so where left and right limit are equal) and jump discontinuity (where it goes from eg. 3 to 9, both not undefined). However I prefer to use the terms over the 'number'

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