The given expression is:

$\lim_{x \to \infty} \left( \sqrt{2x^2 + 3} - \sqrt{2x^2 - 5} \right)$

Step-by-step solution:

1. Factor out $x^2$ under the square roots: For large $x$, both expressions inside the square roots are dominated by $2x^2$. Let's factor $x^2$ out of each square root:

   \sqrt{2x^2 + 3} = \sqrt{x^2(2 + \frac{3}{x^2})} = |x| \sqrt{2 + \frac{3}{x^2}} = x\sqrt{2 + \frac{3}{x^2}} \quad \text{(for large x$,$x > 0)} $\sqrt{2x^2 - 5} = \sqrt{x^2(2 - \frac{5}{x^2})} = x\sqrt{2 - \frac{5}{x^2}}$

2. Rewrite the limit: Now substitute these into the original limit expression: $\lim_{x \to \infty} \left( x\sqrt{2 + \frac{3}{x^2}} - x\sqrt{2 - \frac{5}{x^2}} \right)$ Factor $x$ out: $= x \lim_{x \to \infty} \left( \sqrt{2 + \frac{3}{x^2}} - \sqrt{2 - \frac{5}{x^2}} \right)$

3. Apply the binomial expansion approximation: For large $x$, use the binomial expansion for the square root, keeping only the first term (since $\frac{3}{x^2}$ and $\frac{5}{x^2}$ are very small): $\sqrt{2 + \frac{3}{x^2}} \approx \sqrt{2} + \frac{3}{2\sqrt{2}x^2}$ $\sqrt{2 - \frac{5}{x^2}} \approx \sqrt{2} - \frac{5}{2\sqrt{2}x^2}$

4. Simplify the difference: Subtract the two approximations: $\sqrt{2 + \frac{3}{x^2}} - \sqrt{2 - \frac{5}{x^2}} \approx \left( \sqrt{2} + \frac{3}{2\sqrt{2}x^2} \right) - \left( \sqrt{2} - \frac{5}{2\sqrt{2}x^2} \right)$ $= \frac{3}{2\sqrt{2}x^2} + \frac{5}{2\sqrt{2}x^2} = \frac{8}{2\sqrt{2}x^2} = \frac{4}{\sqrt{2}x^2} = \frac{2\sqrt{2}}{x^2}$

5. Take the limit: Now substitute this back into the limit expression: $\lim_{x \to \infty} x \cdot \frac{2\sqrt{2}}{x^2} = \lim_{x \to \infty} \frac{2\sqrt{2}}{x} = 0$

Final Answer:

$\lim_{x \to \infty} \left( \sqrt{2x^2 + 3} - \sqrt{2x^2 - 5} \right) = 0$

Would you like further details on any of the steps or have any questions?

Here are 5 related questions to explore:

1. How does the binomial expansion help in simplifying limits with square roots?
2. What happens if the constants inside the square roots are negative?
3. How can this method be applied to other limits involving square roots?
4. What are some other techniques for evaluating limits at infinity?
5. How does this limit change if the terms inside the square roots are multiplied by different powers of $x$?

Tip: When simplifying limits at infinity with square roots, factoring out the dominant term often makes calculations easier.