$\lim_{x\to \infty} \sqrt{x^2+9} - \sqrt{x^2-2}$

I have tried multiplying by the conjugate but the square roots are throwing me off and I'm not sure what to do next. How do you solve this?

  • $\begingroup$ What do you get after you multiply by the conjugate? $\endgroup$ – Joe Johnson 126 Jul 9 '14 at 21:18
  • $\begingroup$ You should get $\frac{11}{(x^2+9)^{1/2}+(x^2-2)^{1/2}}$ after conjugating $\endgroup$ – illysial Jul 9 '14 at 21:19
  • $\begingroup$ Indeed I do get exactly what @illysial posted, except I left the root symbols. $\endgroup$ – jrounsav Jul 9 '14 at 21:22
  • $\begingroup$ Do you see what the denominator approaches? $\endgroup$ – illysial Jul 9 '14 at 21:22
  • $\begingroup$ Now it is clear that as $x$ gets big, the thing dies. $\endgroup$ – André Nicolas Jul 9 '14 at 21:22


$$\lim_{x \to \infty} \sqrt{x^2+9}-\sqrt{x^2-2}=\lim_{x \to \infty } \frac{11}{\sqrt{x^2+9}+\sqrt{x^2-2}}=0$$


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