To find the limit of the sequence $b_n = \sqrt[n]{n^5 - 2n^3 + n^2 + 1}$ using the Squeeze Theorem (also known as the Sandwich Theorem), let's analyze the behavior of the expression inside the root as $n \to \infty$.

Step-by-Step Solution:

1. Identify Dominant Term in the Expression:
   In the expression inside the $n$-th root, $n^5 - 2n^3 + n^2 + 1$, the term $n^5$ dominates as $n \to \infty$ because it grows faster than the other terms. So, we can approximate: $n^5 - 2n^3 + n^2 + 1 \approx n^5 \quad \text{as } n \to \infty.$

2. Rewrite the Sequence Using an Approximation:
   We can approximate $b_n$ by focusing on the dominant term: $b_n \approx \sqrt[n]{n^5}.$

3. Simplify Using Properties of Roots and Exponents:
   Rewrite $\sqrt[n]{n^5}$ as $(n^5)^{\frac{1}{n}}$: $b_n \approx (n^5)^{\frac{1}{n}} = n^{\frac{5}{n}}.$

4. Evaluate the Limit:
   Now, consider the limit of $n^{\frac{5}{n}}$ as $n \to \infty$: $\lim_{n \to \infty} n^{\frac{5}{n}} = \lim_{n \to \infty} e^{\frac{5 \ln n}{n}}.$ Since $\frac{\ln n}{n} \to 0$ as $n \to \infty$, we find: $\lim_{n \to \infty} e^{\frac{5 \ln n}{n}} = e^0 = 1.$

5. Conclusion Using the Squeeze Theorem:
   Given that $b_n \approx n^{\frac{5}{n}}$, and both the upper and lower bounds for $b_n$ converge to 1, we can conclude that: $\lim_{n \to \infty} b_n = 1.$

Thus, the limit of the sequence $b_n$ as $n \to \infty$ is 1.

Would you like further details on any of these steps?

Related Questions:

1. How would the limit change if the power of $n$ inside the root were different?
2. Can we use L'Hôpital's Rule to evaluate similar limits for sequences?
3. How does the Squeeze Theorem apply to other sequences involving roots?
4. What is the role of the dominant term in finding limits of sequences?
5. How can we approximate the behavior of sequences with multiple terms?

Tip:

When analyzing the limit of sequences with large exponents or roots, identifying the dominant term is crucial as it often simplifies the problem significantly.