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lim_{x rightarrow 1-} (sqrt{2x}\left| 1-x \right|)div(x ^{2}-1)
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\[\lim_{x \rightarrow 1-} (\sqrt{2x}\left| 1-x \right|)\div(x ^{2}-1) \]
ya
For 0<x<1 we have \[\frac{\sqrt{2x}|1-x|}{x^2-1}=-\frac{\sqrt{2x}(1-x)}{(x-1)(x+1)} = - \frac{\sqrt{2x}}{x+1} \] Now you can insert x=1, so the limit is \[-\sqrt{2}/2\]
Oh sorry, erase the sign after the first equality! But that doesn't change anything.
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