# Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 34

Determine the $\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+2}}{\sqrt{2x^2+1}}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

$\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+2}}{\sqrt{2x^2+1}} = \frac{\sqrt{\infty^2 + 2}}{\sqrt{2(\infty)^2+1}} = \frac{\sqrt{\infty}}{\sqrt{\infty}} = \frac{\infty}{\infty} \text{ Indeterminate}$

Thus, by applying L'Hospital's Rule...

\begin{aligned} \lim_{x \to \infty} \frac{\sqrt{x^2+2}}{\sqrt{2x^2+1}} &= \frac{\frac{2x}{2\sqrt{x^2+2}}}{\frac{4x}{2\sqrt{2x^2+1}}}\\ \\ &= \lim_{x \to \infty} \frac{\sqrt{2x^2+1}}{2\sqrt{x^2+2}} \end{aligned}

Again, if we apply L'Hospital's Rule...

\begin{aligned} \lim_{x \to \infty} \frac{\sqrt{2x^2+1}}{2\sqrt{x^2+2}} &= \lim_{x \to \infty} \frac{\frac{4x}{2\sqrt{2x^2 +1}}}{2 \cdot \frac{2x}{2\sqrt{x^2+2}}}\\ \\ &= \lim_{x \to \infty} \frac{\sqrt{x^2+2}}{\sqrt{2x^2 +1}} \end{aligned}

Notice that we can't apply L'Hospital's Rule since we can't simplify the function and eliminate the square root sign.

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