The expression provided in the image is:

$\lim_{{x \to 1}} \frac{x^{100} - 2x + 1}{x^{50} - 2x + 1}$

Steps to solve:

1. Substitute $x = 1$ into the expression:

   $\frac{1^{100} - 2(1) + 1}{1^{50} - 2(1) + 1} = \frac{1 - 2 + 1}{1 - 2 + 1} = \frac{0}{0}$

   This results in an indeterminate form $\frac{0}{0}$, which means we need to apply L'Hopital's Rule to find the limit.

2. Apply L'Hopital's Rule: This rule states that if we encounter an indeterminate form, we can differentiate the numerator and the denominator separately with respect to $x$, then recompute the limit.

   • The derivative of the numerator $x^{100} - 2x + 1$ is:

     $\frac{d}{dx}(x^{100} - 2x + 1) = 100x^{99} - 2$

   • The derivative of the denominator $x^{50} - 2x + 1$ is:

     $\frac{d}{dx}(x^{50} - 2x + 1) = 50x^{49} - 2$

3. Recompute the limit using these derivatives:

   $\lim_{{x \to 1}} \frac{100x^{99} - 2}{50x^{49} - 2}$

   Substitute $x = 1$:

   $\frac{100(1)^{99} - 2}{50(1)^{49} - 2} = \frac{100 - 2}{50 - 2} = \frac{98}{48} = \frac{49}{24}$

Final Answer:

$\lim_{{x \to 1}} \frac{x^{100} - 2x + 1}{x^{50} - 2x + 1} = \frac{49}{24}$

Would you like more details or have any questions?

Here are 5 related questions to practice:

1. How do you apply L'Hopital's Rule to limits involving higher powers?
2. What are the conditions required to apply L'Hopital's Rule?
3. How can you simplify limits of rational functions without using L'Hopital's Rule?
4. What happens if L'Hopital's Rule gives another indeterminate form?
5. Can the quotient of polynomials always be solved using L'Hopital's Rule?

Tip: Always check if the limit results in an indeterminate form before applying L'Hopital's Rule.