How do I find lim n → ∞ ( n 3 + 1 n 3 ) 1 n lim � → ∞ ( � 3 + 1 � 3 ) 1 � ?

 To find the limit of the expression lim n→∞(n3+1n3)1n, you can use various methods depending on your preferred approach:

Method 1: L'Hôpital's Rule:

1. Recognize the indeterminate form 1∞. This form arises when both the numerator and denominator approach infinity as n approaches infinity.

2. Apply L'Hôpital's rule repeatedly until you reach a non-indeterminate form. This rule allows you to differentiate both the numerator and denominator as long as the resulting limit is finite and defined.

3. Differentiate the numerator and denominator once: lim n→∞(3n2−3n−1)n2

4. Evaluate the new limit directly or use L'Hôpital's rule again if needed. In this case, the limit becomes 0, which is the answer.

Method 2: Simplifying with Algebraic Manipulation:

1. Notice that the expression involves n^3 in both the numerator and denominator. You can simplify by dividing both by n^3:

lim n→∞(1+1n2)1n

1. As n approaches infinity, 1n2 approaches 0, so you can replace it with 0 for simplification:

lim n→∞(1+0)1n = lim n→∞11n = 1

Method 3: Squeeze Theorem:

1. Find two functions, g(x) and h(x), such that for all n > N (some large value):

g(n) ≤ (n3+1n3)1n ≤ h(n)

2. Ensure g(n) and h(n) both have a limit of 1 as n approaches infinity. For example, you can choose g(n) = 1 - 1/n and h(n) = 1 + 1/n.

3. Apply the squeeze theorem: If for all x > N, g(x) ≤ f(x) ≤ h(x), and both g(x) and h(x) approach the same limit L as x approaches infinity, then f(x) also approaches L as x approaches infinity.

In this case, applying the squeeze theorem with the chosen g(n) and h(n) confirms the limit of the original expression is 1.

All three methods lead to the same answer: lim n→∞(n3+1n3)1n = 1.

Choose the method that best suits your understanding and preference for solving such limits.