# How do you find the following limit as x approaches infinity?

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

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?

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

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