Sequence Analysis: Inf, Sup, Lim Inf, And Lim Sup Explained

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Hey math enthusiasts! Let's dive into the fascinating world of sequences and explore some key concepts: infimum, supremum, limit inferior, and limit superior. These terms help us understand the behavior of sequences, especially when they don't behave nicely, like when they don't converge. We'll be working through two examples, breaking down each step to make sure everything clicks. Ready to get started? Let's go!

99. Analyzing the Sequence xn=n2−9n−100x_n = n^2 - 9n - 100

Alright, guys, let's get our hands dirty with the first sequence: xn=n2−9n−100x_n = n^2 - 9n - 100. Our goal is to find its infimum, supremum, limit inferior, and limit superior. These concepts give us a complete picture of the sequence's behavior, especially its bounds and long-term trends. So, buckle up; we're about to analyze this sequence thoroughly!

First, let's understand what each of these terms means in plain English. The infimum is the greatest lower bound – the largest number that is less than or equal to all terms in the sequence. The supremum is the least upper bound – the smallest number that is greater than or equal to all terms. The limit inferior is the smallest limit point – the smallest value that the sequence approaches infinitely often. The limit superior is the largest limit point – the largest value the sequence approaches infinitely often. Got it? Cool!

To begin, let's explore the behavior of xn=n2−9n−100x_n = n^2 - 9n - 100 as n increases. This is a quadratic function, and we know that quadratics have a parabolic shape. Because the coefficient of n2n^2 is positive (it's 1), the parabola opens upwards. This means the sequence will eventually increase without bound as n grows. Initially, though, it might decrease.

Let's find the vertex of the parabola, which corresponds to the minimum value of the quadratic function. The x-coordinate (in this case, n-coordinate) of the vertex can be found using the formula n = -b / 2a, where a and b are the coefficients of the quadratic equation. Here, a = 1 and b = -9. Therefore, the vertex's n-coordinate is n = -(-9) / (2 * 1) = 4.5. Since n must be a positive integer, we'll consider the terms around n = 4 and n = 5. Let's calculate a few terms:

  • x4=42−9(4)−100=16−36−100=−120x_4 = 4^2 - 9(4) - 100 = 16 - 36 - 100 = -120
  • x5=52−9(5)−100=25−45−100=−120x_5 = 5^2 - 9(5) - 100 = 25 - 45 - 100 = -120
  • x6=62−9(6)−100=36−54−100=−118x_6 = 6^2 - 9(6) - 100 = 36 - 54 - 100 = -118

Notice that the value of the sequence reaches a minimum near n=4 or n=5. We can calculate the value at n = 4.5, which is the exact vertex, though we'll need to round this to the nearest integers because n is an integer. Thus, the minimum value can be understood through the values of the function close to 4.5. As n increases from 5, the sequence starts to increase. This tells us a lot about the infimum, supremum, and the limits!

Now, let's identify the infimum. Since the parabola opens upwards and has a minimum value, there is a greatest lower bound. Analyzing the first few terms, it's clear the sequence takes on values as small as -120 and then begins to increase. Since the parabola continues to increase to infinity, we can conclude the infimum is the minimum value. Since the term is an integer and the vertex is 4.5, we can consider the two adjacent points which give the same value. So, looking at the terms, the greatest lower bound, or the infimum, is -120.

Next, let's talk about the supremum. Because the parabola opens upwards, the sequence increases without bound. It doesn't have an upper bound, so the supremum is positive infinity (+∞). Any value greater than some point will eventually be smaller than the subsequent value of the sequence.

Regarding the limit inferior, since the sequence approaches the minimum value (-120) and then grows to infinity, we can say that the limit inferior is -120. This is because the sequence approaches -120 infinitely often (at least once) from below, and it is the smallest limit point.

Finally, for the limit superior, the sequence increases without bound, approaching positive infinity (+∞). Therefore, the limit superior is +∞. This is the largest value the sequence approaches infinitely often.

In summary, for the sequence xn=n2−9n−100x_n = n^2 - 9n - 100:

  • Infimum: -120
  • Supremum: +∞
  • Limit Inferior: -120
  • Limit Superior: +∞

We've successfully analyzed the first sequence, guys! It may seem tricky, but by taking it one step at a time, we can understand the behavior of the sequence.

100. Analyzing the Sequence x_n = n + rac{100}{n}

Alright, let's move on to the second sequence: x_n = n + rac{100}{n}. This one might seem a bit different, but the principles remain the same. We'll find the infimum, supremum, limit inferior, and limit superior. Again, these values describe the sequence's bounds and long-term behavior. Are you ready?

Let's start by understanding how this sequence behaves as n increases. The term n increases without bound as n grows. However, the term rac{100}{n} decreases as n increases. The overall behavior of the sequence is determined by a balance of these two terms. Because the n term grows and dominates, we can see the sequence will also increase without bound.

To find the infimum, we need to analyze the initial behavior of the sequence. For small values of n, the term rac{100}{n} will have a more significant impact. Let's calculate a few terms to get a better idea:

  • x_1 = 1 + rac{100}{1} = 101
  • x_2 = 2 + rac{100}{2} = 52
  • x_3 = 3 + rac{100}{3} hickapprox 36.33
  • x_4 = 4 + rac{100}{4} = 29
  • x_5 = 5 + rac{100}{5} = 25
  • x_6 = 6 + rac{100}{6} hickapprox 22.67
  • x_7 = 7 + rac{100}{7} hickapprox 21.29
  • x_8 = 8 + rac{100}{8} = 20.5
  • x_9 = 9 + rac{100}{9} hickapprox 20.11
  • x_{10} = 10 + rac{100}{10} = 20
  • x_{11} = 11 + rac{100}{11} hickapprox 20.09
  • x_{12} = 12 + rac{100}{12} hickapprox 20.33

Notice that the sequence decreases initially, reaches a minimum point, and then starts to increase. The minimum appears to be at around n = 10.

To find the exact minimum, consider the derivative of the continuous function f(x) = x + rac{100}{x}. The derivative is f'(x) = 1 - rac{100}{x^2}. Setting this to zero to find the critical points, we get 1 = rac{100}{x^2}, or x2=100x^2 = 100. This gives us x=10x = 10 and x=−10x = -10. Since we are working with a sequence where n is positive, we focus on x=10. The function's value at this point is f(10) = 10 + rac{100}{10} = 20. Since we have an integer sequence, the minimum is 20, as our calculations showed.

So, the greatest lower bound, or the infimum, is 20.

Now, for the supremum. As n grows, the term n dominates the sequence, and the sequence increases without bound. The supremum is thus positive infinity (+∞).

For the limit inferior, we analyze the smallest value the sequence approaches infinitely often. The sequence approaches 20 as n gets close to 10 and then moves away. Thus, the limit inferior is 20.

Finally, for the limit superior, the sequence approaches positive infinity (+∞) as n grows. The limit superior is +∞.

In summary, for the sequence x_n = n + rac{100}{n}:

  • Infimum: 20
  • Supremum: +∞
  • Limit Inferior: 20
  • Limit Superior: +∞

And there you have it, folks! We've successfully analyzed the second sequence, uncovering its infimum, supremum, limit inferior, and limit superior. Good job!

I hope this explanation has been helpful. Keep practicing, and you'll become a sequence analysis pro! If you want, you can try with other examples, and I can walk you through it too. Math is all about practice, and hopefully this explanation makes it a lot easier and less intimidating. Happy calculating!