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 December 13th, 2017, 10:27 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 Show that... $\sqrt{x^{2}+a^{2}}-x$ approaches 0 as x approaches infinity
 December 13th, 2017, 10:42 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics 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, $\sqrt{x^2 + a^2} + x \to \infty$. Thanks from Jaket1
 December 13th, 2017, 10:56 AM #3 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 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.
December 13th, 2017, 10:59 AM   #4
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Joined: Sep 2016
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Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by Jaket1 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.

 December 14th, 2017, 01:56 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond $$\left(\sqrt{x^2-a^2}-x\right)\cdot\frac{\sqrt{x^2-a^2}+x}{\sqrt{x^2-a^2}+x}=-\frac{a^2}{\sqrt{x^2-a^2}+x}$$

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