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 December 13th, 2017, 11:27 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, 11:42 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 685 Thanks: 461 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, 11:56 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, 11:59 AM   #4
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 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, 02:56 AM #5 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,981 Thanks: 1166 Math Focus: Elementary mathematics and beyond $$\left(\sqrt{x^2-a^2}-x\right)\cdot\frac{\sqrt{x^2-a^2}+x}{\sqrt{x^2-a^2}+x}=-\frac{a^2}{\sqrt{x^2-a^2}+x}$$ Tags show Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fahad nasir Calculus 1 July 24th, 2016 05:13 AM mared Trigonometry 5 March 26th, 2015 07:19 PM notnaeem Real Analysis 4 August 16th, 2010 01:32 PM naserellid Algebra 0 August 15th, 2010 11:25 AM naserellid Algebra 2 August 15th, 2010 03:20 AM

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