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>> No.15800922 [View]
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15800922

If there is a limit going towards some value [math]a^2[/math] such that [math]a>0[/math] is in the argument's denominator [math]\sqrt{x}+a[/math], why do we ignore the negative root of [math]a^2[/math] here and just proceed to let [math]x=a^2[/math] and get rid of the limit?

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