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December 13th, 2017, 10:27 AM   #1
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\[\sqrt{x^{2}+a^{2}}-x\] approaches 0 as x approaches infinity
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December 13th, 2017, 10:42 AM   #2
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Hint: $a^2 = x^2 + a^2 - x^2 = (\sqrt{x^2+a^2} - x)(\sqrt{x^2 + a^2}+x)$ is bounded for all $x$. However, $\sqrt{x^2 + a^2} + x \to \infty$.
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December 13th, 2017, 10:56 AM   #3
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Does this mean that some sequence for x tending to infinity multiplied by another sequence= a finite number if and only if the other sequence approaches 0? That sequence being the one we are asked to prove? Thanks.
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December 13th, 2017, 10:59 AM   #4
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Does this mean that some sequence for x tending to infinity multiplied by another sequence= a finite number if and only if the other sequence approaches 0? That sequence being the one we are asked to prove? Thanks.
You got it.
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December 14th, 2017, 01:56 AM   #5
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$$\left(\sqrt{x^2-a^2}-x\right)\cdot\frac{\sqrt{x^2-a^2}+x}{\sqrt{x^2-a^2}+x}=-\frac{a^2}{\sqrt{x^2-a^2}+x}$$
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