Limit Square Roots E3E03D

1. **State the problem:** We want to find the limit $$\lim_{x \to \pm \infty} \left( \sqrt{3x^2+1} - \sqrt{2x^2+x} \right).$$ 2. **Recall the formula and technique:** When dealing with limits involving differences of square roots, a useful technique is to multiply and divide by the conjugate to simplify the expression: $$\sqrt{a} - \sqrt{b} = \frac{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})}{\sqrt{a} + \sqrt{b}} = \frac{a - b}{\sqrt{a} + \sqrt{b}}.$$ 3. **Apply the conjugate:** Multiply numerator and denominator by $$\sqrt{3x^2+1} + \sqrt{2x^2+x}$$: $$\lim_{x \to \pm \infty} \frac{(3x^2+1) - (2x^2+x)}{\sqrt{3x^2+1} + \sqrt{2x^2+x}} = \lim_{x \to \pm \infty} \frac{3x^2 + 1 - 2x^2 - x}{\sqrt{3x^2+1} + \sqrt{2x^2+x}} = \lim_{x \to \pm \infty} \frac{x^2 - x + 1}{\sqrt{3x^2+1} + \sqrt{2x^2+x}}.$$ 4. **Factor out $$x^2$$ in numerator and denominator terms inside the roots to simplify:** $$\sqrt{3x^2+1} = \sqrt{x^2(3 + \frac{1}{x^2})} = |x| \sqrt{3 + \frac{1}{x^2}},$$ $$\sqrt{2x^2+x} = \sqrt{x^2(2 + \frac{1}{x})} = |x| \sqrt{2 + \frac{1}{x}}.$$ 5. **Rewrite the limit expression:** $$\lim_{x \to \pm \infty} \frac{x^2 - x + 1}{|x| \left( \sqrt{3 + \frac{1}{x^2}} + \sqrt{2 + \frac{1}{x}} \right)}.$$ 6. **Divide numerator and denominator by $$|x|$$:** $$\lim_{x \to \pm \infty} \frac{\frac{x^2}{|x|} - \frac{x}{|x|} + \frac{1}{|x|}}{\sqrt{3 + \frac{1}{x^2}} + \sqrt{2 + \frac{1}{x}}}.$$ 7. **Simplify $$\frac{x^2}{|x|} = |x|$$ and $$\frac{x}{|x|} = \pm 1$$ depending on the sign of $$x$$:** - For $$x \to +\infty$$, $$\frac{x}{|x|} = 1$$ and $$|x| = x$$. - For $$x \to -\infty$$, $$\frac{x}{|x|} = -1$$ and $$|x| = -x$$. 8. **Evaluate the limit as $$x \to +\infty$$:** $$\lim_{x \to +\infty} \frac{x - 1 + 0}{\sqrt{3 + 0} + \sqrt{2 + 0}} = \lim_{x \to +\infty} \frac{x - 1}{\sqrt{3} + \sqrt{2}} = +\infty.$$ 9. **Evaluate the limit as $$x \to -\infty$$:** $$\lim_{x \to -\infty} \frac{-x - (-1) + 0}{\sqrt{3 + 0} + \sqrt{2 + 0}} = \lim_{x \to -\infty} \frac{-x + 1}{\sqrt{3} + \sqrt{2}}.$$ Since $$x \to -\infty$$, $$-x \to +\infty$$, so the numerator tends to $$+\infty$$ and denominator is positive, so the limit is $$+\infty$$. **Final answer:** $$\lim_{x \to \pm \infty} \left( \sqrt{3x^2+1} - \sqrt{2x^2+x} \right) = +\infty.$$