Why does the limit of 1/x² not exist as x approaches zero?

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Homework Statement


Show limit 1/x^2 when x approaches zero does not exist.


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The Attempt at a Solution


What I do is suppose the limit exists,say L.Then I show that for all real number L, the limit does not approaches L. In this case, I separate L into different cases. Can anyone help me to check is this a valid prove.
 
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Yes, that is a valid approach.
 
\lim_{x \to 0}\frac{1}{x^2} = \infty

Since ∞ is not a real number the limit above is said not to exist. This limit means that for any large, positive number M, there is some interval around 0 such that if x is in that interval, 1/x2 > M.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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