To evaluate the limit

$\lim_{x \to \infty} \frac{\sqrt{8x^2 + 7}}{3 - 2x},$

we can analyze the behavior of the numerator and the denominator as $x$ approaches infinity.

Step-by-Step Solution

1. Identify the dominant terms:

   • As $x \to \infty$, the dominant term in the numerator $\sqrt{8x^2 + 7}$ is $\sqrt{8x^2} = 2\sqrt{2}x$.
   • The dominant term in the denominator $3 - 2x$ is $-2x$.

2. Simplify the expression: Divide both the numerator and the denominator by $x$ to simplify:

   $\frac{\sqrt{8x^2 + 7}}{3 - 2x} = \frac{\frac{\sqrt{8x^2 + 7}}{x}}{\frac{3}{x} - 2}.$

3. Simplify further:

   • For the numerator: $\frac{\sqrt{8x^2 + 7}}{x} = \sqrt{8 + \frac{7}{x^2}}$.
   • For the denominator: $\frac{3}{x} - 2 \to -2$ as $x \to \infty$.

   So the expression simplifies to:

   $\frac{\sqrt{8 + \frac{7}{x^2}}}{-2}.$

4. Take the limit as $x \to \infty$: As $x \to \infty$, $\frac{7}{x^2} \to 0$, so the expression inside the square root becomes:

   $\sqrt{8 + 0} = \sqrt{8} = 2\sqrt{2}.$

   Therefore, the limit is:

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

Final Answer

$\lim_{x \to \infty} \frac{\sqrt{8x^2 + 7}}{3 - 2x} = -\sqrt{2}.$

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

Related Questions

1. How would the limit change if the denominator was $3 + 2x$?
2. What happens if the numerator was $\sqrt{8x^2 - 7}$ instead of $\sqrt{8x^2 + 7}$?
3. Can you find the limit as $x \to -\infty$ for the same expression?
4. How does the presence of the square root affect the rate at which the numerator grows compared to the denominator?
5. How would you evaluate $\lim_{x \to \infty} \frac{x^2 + 7}{3 - 2x}$?

Tip

When taking limits as $x \to \infty$, dividing all terms by the highest power of $x$ in the denominator often helps to simplify the expression and identify the behavior of the limit.