1
$\begingroup$

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

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

$$\sqrt{x^2+9}-\sqrt{x^2-2}=\frac{(\sqrt{x^2+9}-\sqrt{x^2-2})(\sqrt{x^2+9}+\sqrt{x^2-2})}{\sqrt{x^2+9}+\sqrt{x^2-2}}=\frac{x^2+9-x^2+2}{\sqrt{x^2+9}+\sqrt{x^2-2}}=\frac{11}{\sqrt{x^2+9}+\sqrt{x^2-2}}$$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.