The problem is asking for the limit of the function as $x$ approaches negative infinity:

$\lim_{x \to -\infty} \frac{\sqrt{2x^2 + 3}}{2x + 3}$

Solution:

1. Simplify the Expression:

   • First, notice that as $x$ becomes very large in magnitude (negatively), the term $2x^2$ dominates over the constant terms $3$ in both the numerator and the denominator.
   • We can factor $x^2$ out of the square root in the numerator.

   $\frac{\sqrt{2x^2 + 3}}{2x + 3} = \frac{\sqrt{x^2(2 + \frac{3}{x^2})}}{2x + 3} = \frac{|x|\sqrt{2 + \frac{3}{x^2}}}{2x + 3}$

2. Consider the Sign of $x$:

   • Since $x$ is approaching $-\infty$, $|x| = -x$. So the expression becomes:

   $\frac{-x\sqrt{2 + \frac{3}{x^2}}}{2x + 3}$

3. Simplify Further:

   • Divide both the numerator and the denominator by $x$:

   $\frac{-\sqrt{2 + \frac{3}{x^2}}}{2 + \frac{3}{x}}$

4. Evaluate the Limit:

   • As $x \to -\infty$, $\frac{3}{x^2}$ and $\frac{3}{x}$ approach 0, simplifying the expression to:

   $\frac{-\sqrt{2}}{2}$

   • Therefore, the limit is:

   $\lim_{x \to -\infty} \frac{\sqrt{2x^2 + 3}}{2x + 3} = -\frac{\sqrt{2}}{2}$

Would you like more details, or do you have any questions about this solution?

Related Questions:

1. How does the behavior of the function change as $x$ approaches positive infinity instead of negative infinity?
2. What is the importance of considering the sign of $x$ when dealing with absolute values in limits?
3. Can you explain why we factor $x^2$ out of the square root in the numerator?
4. How would the limit change if the constant in the numerator was a different value?
5. What happens to the limit if the term $\sqrt{2x^2 + 3}$ was replaced with $\sqrt{x^4 + 3}$?

Tip:

When dealing with limits involving infinity, simplifying the expression by factoring out the highest power of $x$ can often make the limit more straightforward to evaluate.