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To find the limit as \( x \) approaches infinity for the expression \(\sqrt{(2x-5)(2x+1)} - (2x-5)\), we can start by simplifying the expression under the square root.

The expression simplifies as follows:
\[
\sqrt{(2x-5)(2x+1)} = \sqrt{4x^2 - 8x - 5}
\]

When \( x \) is very large, the term \( -8x \) becomes negligible in comparison to \( 4x^2 \), so we can approximate this as:
\[
\sqrt{4x^2 - 8x - 5} \approx \sqrt{4x^2} = 2x
\]

Substituting this back into the original expression, we get:
\[
\sqrt{(2x-5)(2x+1)} - (2x-5) \approx 2x - (2x-5) = 5
\]

However, this result seems inconsistent with the choices provided, indicating a mistake in my initial assumption or simplification. Let's re-evaluate the expression more rigorously:

Consider the expression as \( x \to \infty \):
\[
\sqrt{(2x-5)(2x+1)} - (2x-5)
\]
\[
= \sqrt{4x^2 - 8x - 5} - (2x-5)
\]

We can use the approximation for square roots involving large numbers:
\[
\sqrt{4x^2 - 8x - 5} \approx 2x - \frac{8x + 5}{4x} = 2x - 2 - \frac{5}{4x}
\]
\[
\approx 2x - 2 \quad \text{(as } x \to \infty)
\]

Thus,
\[
\sqrt{(2x

Find the limit of the expression \(\lim_{{x \to \infty}} \sqrt{{(2x - 5)(2x + 1)}} - (2x - 5)\).

This problem involves finding the limit of the expression \(\sqrt{{(2x - 5)(2x + 1)}} - (2x - 5)\) as x approaches infinity. Learn how to approximate square roots and simplify limits for large x values. Ideal for high school students in Grades 10-12.

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