It seems like there is a formatting issue with the mathematical expression. Based on what you provided, it appears you're asking to compute the limit of a function as $x \to -\infty$. Here's how I interpret your expression:

$\lim_{{x \to -\infty}} \sqrt{\frac{x^2 - 4}{2x + 5}}$

Let me solve this step by step:

Step 1: Simplifying the expression inside the square root

We have:

$\frac{x^2 - 4}{2x + 5}$

As $x \to -\infty$, the term $x^2$ dominates over the constant $-4$ in the numerator. So:

$x^2 - 4 \approx x^2$

Step 2: Handling the denominator

In the denominator, as $x \to -\infty$, the linear term $2x$ dominates over the constant $5$. So:

$2x + 5 \approx 2x$

Step 3: Simplifying the fraction

Now, we approximate the fraction:

$\frac{x^2}{2x} = \frac{x}{2}$

Step 4: Applying the square root

We now take the square root of the simplified expression:

$\sqrt{\frac{x}{2}} = \frac{\sqrt{x}}{\sqrt{2}}$

However, since $x \to -\infty$, we have to consider the fact that $\sqrt{x}$ is undefined for negative numbers. But, we can express $\sqrt{x^2} = |x|$, and since $x \to -\infty$, $|x| = -x$. Therefore:

$\sqrt{\frac{x^2}{2x}} = \frac{-x}{\sqrt{2}}$

Step 5: Final answer

Thus, as $x \to -\infty$, the limit becomes:

$\lim_{{x \to -\infty}} \sqrt{\frac{x^2 - 4}{2x + 5}} = \frac{-x}{\sqrt{2}} \to +\infty$

So, the limit is $+\infty$.

Would you like further clarification or details on any step?

Here are some related questions:

1. How does one handle square roots in limits approaching negative infinity?
2. What role does dominance of terms play in simplifying limits?
3. Can limits approach infinity in functions with square roots?
4. What happens if the expression inside the square root is negative?
5. How does the sign of $x$ affect the final result?

Tip: Always check whether the terms inside square roots are positive or negative, especially when approaching negative infinity.