The problem in the image is asking to evaluate the limit:

$\lim_{x \to \infty} \left( 1 + \frac{5}{2x^2 - 3x} \right)^{4x^2}.$

This is a classic case involving limits of the exponential type, where the base approaches $1$ and the exponent approaches infinity. Let us carefully evaluate this step by step.

Step 1: Rewrite the expression

Set: $y = \left( 1 + \frac{5}{2x^2 - 3x} \right)^{4x^2}.$ Take the natural logarithm: $\ln y = 4x^2 \ln \left( 1 + \frac{5}{2x^2 - 3x} \right).$

Step 2: Approximation for $\ln(1 + u)$

For small $u$, $\ln(1 + u) \approx u$. Here, as $x \to \infty$, $\frac{5}{2x^2 - 3x} \to 0$. So, we can approximate: $\ln \left( 1 + \frac{5}{2x^2 - 3x} \right) \approx \frac{5}{2x^2 - 3x}.$

Substitute this back into the expression for $\ln y$: $\ln y \approx 4x^2 \cdot \frac{5}{2x^2 - 3x}.$

Step 3: Simplify the expression

Factor $x^2$ from the denominator: $\frac{5}{2x^2 - 3x} = \frac{5}{x^2 \left( 2 - \frac{3}{x} \right)}.$

As $x \to \infty$, the term $\frac{3}{x} \to 0$, so: $\frac{5}{2x^2 - 3x} \to \frac{5}{2x^2}.$

Thus: $\ln y \approx 4x^2 \cdot \frac{5}{2x^2} = \frac{20}{2} = 10.$

Step 4: Exponentiate to find $y$

Since $\ln y \to 10$, we have: $y \to e^{10}.$

Final Answer:

$\lim_{x \to \infty} \left( 1 + \frac{5}{2x^2 - 3x} \right)^{4x^2} = e^{10}.$

Let me know if you'd like additional clarifications or further explanations!

Follow-Up Questions:

1. How is the approximation $\ln(1 + u) \approx u$ derived for small $u$?
2. Why is it valid to ignore higher-order terms in the denominator for large $x$?
3. What happens if the exponent $4x^2$ were replaced with a smaller growth rate (e.g., $x$)?
4. Could this problem be solved using L’Hôpital’s Rule? Why or why not?
5. How can you generalize this approach for limits of the form $\lim_{x \to \infty} \left(1 + \frac{a}{b(x)}\right)^{c(x)}$?

Tip:

When evaluating limits involving exponential forms, always check if the base approaches $1$ and the exponent grows without bounds—this indicates the use of logarithms and approximations!