This is a mathematical limit problem. The goal is to find:

$\lim_{x \to \infty} \frac{-7x}{\sqrt{4x^2 - 3}}$

Step-by-step Solution:

1. Factor out $x^2$ from the denominator: In the square root, factor $x^2$ from the terms inside: $\sqrt{4x^2 - 3} = \sqrt{x^2 (4 - \frac{3}{x^2})} = |x| \sqrt{4 - \frac{3}{x^2}}$ For $x \to \infty$, $|x| = x$ (since $x > 0$).

   So: $\sqrt{4x^2 - 3} = x \sqrt{4 - \frac{3}{x^2}}$

2. Rewrite the limit: Substitute the simplified denominator back into the fraction: $\frac{-7x}{\sqrt{4x^2 - 3}} = \frac{-7x}{x \sqrt{4 - \frac{3}{x^2}}} = \frac{-7}{\sqrt{4 - \frac{3}{x^2}}}$

3. Simplify as $x \to \infty$: As $x \to \infty$, the term $\frac{3}{x^2} \to 0$. Therefore: $\sqrt{4 - \frac{3}{x^2}} \to \sqrt{4} = 2$

   So, the fraction becomes: $\frac{-7}{\sqrt{4 - \frac{3}{x^2}}} \to \frac{-7}{2}$

Final Answer:

$\lim_{x \to \infty} \frac{-7x}{\sqrt{4x^2 - 3}} = -\frac{7}{2}$

Would you like more details or have any questions?

Related Questions:

1. What happens to the limit if the numerator is $7x$ instead of $-7x$?
2. How does factoring help simplify expressions involving square roots?
3. Why does $|x| = x$ for $x > 0$?
4. How can we handle limits involving $\sqrt{x^2 + a}$ as $x \to \infty$?
5. Can L'Hôpital's Rule be applied to this problem? Why or why not?

Tip:

Always check if terms like $\sqrt{x^2 - a}$ can be simplified by factoring out $x^2$, as it often simplifies limits involving infinity.