December 13th, 2017, 11:27 AM  #1 
Senior Member Joined: Jan 2016 From: Blackpool Posts: 100 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: 531 Thanks: 304 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, $\sqrt{x^2 + a^2} + x \to \infty$.

December 13th, 2017, 11:56 AM  #3 
Senior Member Joined: Jan 2016 From: Blackpool Posts: 100 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 
Senior Member Joined: Sep 2016 From: USA Posts: 531 Thanks: 304 Math Focus: Dynamical systems, analytic function theory, numerics  
December 14th, 2017, 02:56 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,898 Thanks: 1093 Math Focus: Elementary mathematics and beyond 
$$\left(\sqrt{x^2a^2}x\right)\cdot\frac{\sqrt{x^2a^2}+x}{\sqrt{x^2a^2}+x}=\frac{a^2}{\sqrt{x^2a^2}+x}$$


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