To evaluate the limit

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

we can start by simplifying the expression under the square root and identifying the dominant terms as $x$ approaches infinity.

Step-by-Step Solution

1. Rewrite the expression: We start by factoring out $x^2$ inside the square root:

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

2. Simplify the expression: The expression now is:

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

   Factor $x$ out from the entire expression:

   $x\left(\frac{\sqrt{8 + \frac{7}{x^2}}}{3} - 2\right).$

3. Take the limit as $x \to \infty$: As $x \to \infty$, $\frac{7}{x^2} \to 0$. So, the square root term $\sqrt{8 + \frac{7}{x^2}}$ approaches $\sqrt{8}$.

   Therefore,

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

4. Substitute this into the limit expression: The limit now simplifies to:

   $\lim_{x \to \infty} x\left(\frac{2\sqrt{2}}{3} - 2\right).$

   Simplify the coefficient:

   $\frac{2\sqrt{2}}{3} - 2 = \frac{2\sqrt{2}}{3} - \frac{6}{3} = \frac{2\sqrt{2} - 6}{3}.$

5. Analyze the coefficient: Since $2\sqrt{2} \approx 2.828$, we have $2\sqrt{2} - 6 < 0$. Thus, the coefficient is negative.

6. Determine the final limit: As $x \to \infty$, the expression $x \left(\frac{2\sqrt{2} - 6}{3}\right)$ goes to $-\infty$ because the coefficient is negative.

Final Answer

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

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

Related Questions

1. How would the limit change if the square root expression was $\sqrt{8x^2 - 7}$?
2. What happens to the limit if the constant $2x$ is replaced by $3x$?
3. Can you calculate the limit if $x \to -\infty$ for the same expression?
4. How does this limit relate to limits involving logarithmic functions?
5. How would you evaluate $\lim_{x \to \infty} \frac{\sqrt{8x^2 + 7} - 2x}{x}$?

Tip

Always consider factoring out the dominant term in the expression when dealing with limits as $x \to \infty$ to simplify the analysis.