It seems that extracting the mathematical expression from the image didn't work as expected. However, from the visual inspection of the image, the limit problem appears to be:

$\lim_{x \to 1} \frac{2x^3 + 3x^2 - 4x - 1}{x^3 - 1}$

Let's solve this limit step by step.

1. Factorize the denominator:
   $x^3 - 1$ can be factored as: $x^3 - 1 = (x - 1)(x^2 + x + 1)$

2. Substitute $x = 1$ directly:
   If we substitute $x = 1$ into the numerator and denominator, we get: $\frac{2(1)^3 + 3(1)^2 - 4(1) - 1}{(1)^3 - 1} = \frac{2 + 3 - 4 - 1}{1 - 1} = \frac{0}{0}$ This gives an indeterminate form $\frac{0}{0}$, so we can use L'Hopital's Rule.

3. L'Hopital's Rule:
   Differentiate the numerator and denominator with respect to $x$.

   • Derivative of the numerator $2x^3 + 3x^2 - 4x - 1$: $\frac{d}{dx}(2x^3 + 3x^2 - 4x - 1) = 6x^2 + 6x - 4$
   • Derivative of the denominator $x^3 - 1$: $\frac{d}{dx}(x^3 - 1) = 3x^2$

   Now, the limit becomes: $\lim_{x \to 1} \frac{6x^2 + 6x - 4}{3x^2}$

4. Substitute $x = 1$ into the new expression: $\frac{6(1)^2 + 6(1) - 4}{3(1)^2} = \frac{6 + 6 - 4}{3} = \frac{8}{3}$

Thus, the value of the limit is $\frac{8}{3}$.

Would you like further details or have any questions about the solution?

Here are some related questions:

1. What is L'Hopital's Rule and when is it applied?
2. How do you factorize cubic polynomials like $x^3 - 1$?
3. Can we solve this limit by factoring the numerator instead of using L'Hopital's Rule?
4. What other methods can solve indeterminate forms like $\frac{0}{0}$?
5. How can you apply limits to more complex rational functions?

Tip: Always check if substituting the limit value directly into a function results in an indeterminate form before using L'Hopital's Rule.