Proof: Limit Of A Scaled Sequence In Real Analysis

Hey everyone! Let's dive into a fundamental concept in real analysis: proving the limit of a sequence when it's multiplied by a constant. Specifically, we're tackling this: Given $\lim_{n \to \infty} s_n = L$, prove that $\lim_{n \to \infty} (c \cdot s_n) = c \cdot L$, where c is any real number. This might seem straightforward, but the proof highlights how limits work and how we can manipulate them. This is a super important building block for understanding more complex ideas in calculus and analysis, so pay close attention!

The Core Concept: Scaling Sequences and Their Limits

Alright, so what's the deal? We're starting with a sequence of real numbers, denoted as $s_n$. This sequence has a limit, L, meaning as n (the index of the term in the sequence) gets infinitely large, the terms $s_n$ get closer and closer to L. We're also given a constant, c, which can be any real number (positive, negative, or zero). The question is: if we multiply every term in the sequence by c, will the limit also be multiplied by c? And the answer, as you probably guessed, is yes! This property is super useful because it lets us pull constants out of limits, simplifying calculations and proofs. Understanding this allows us to solve problems in a more simple way. This is because we can just focus on finding the original limit of the sequence. Then all we have to do is multiply the limit by the constant provided to us. Let's not waste any more time, and get to the proof, shall we? Hold on tight!

Now, let's break down the proof step by step to see how it all comes together. This proof utilizes the formal definition of a limit, which is the foundation for everything we do in real analysis. Let's get into it.

Understanding the Formal Definition of a Limit

Before we jump into the proof, let's quickly refresh our memory on the formal definition of a limit. The definition states that $\lim_{n \to \infty} s_n = L$ if and only if, for every $\epsilon > 0$, there exists an integer $N$ such that for all $n > N$, we have $|s_n - L| < \epsilon$. What does this mouthful mean? Basically, for any tiny positive number ($\epsilon$), we can always find a point in the sequence (represented by N) such that all terms after that point are within a distance of $\epsilon$ from the limit L. In simpler terms, we can always find a place in the sequence where the terms get arbitrarily close to L.

Keep in mind that this N depends on the choice of $\epsilon$. The smaller we make $\epsilon$, the larger N might need to be. But no matter how small $\epsilon$ is, N always exists because that is the meaning of the limit. This understanding is key to following the proof. Now that we are clear on the formal definition of the limit, let's get to the proof!

The Proof: Unveiling the Relationship Between Limits and Scalar Multiplication

Okay, let's prove $\lim_{n \to \infty} (c \cdot s_n) = c \cdot L$. We're going to use the formal definition of a limit. Here's how it goes:

1. The Setup: We're given that $\lim_{n \to \infty} s_n = L$. By the definition of a limit, for any $\epsilon > 0$, there exists an $N_1$ such that if $n > N_1$, then $|s_n - L| < \epsilon$. This is the base of our proof.
2. Considering the Constant: We want to show that $\lim_{n \to \infty} (c \cdot s_n) = c \cdot L$. So, we need to show that for any $\epsilon > 0$, there exists an N such that if $n > N$, then $|c \cdot s_n - c \cdot L| < \epsilon$.
3. Case 1: c = 0: If c = 0, then $c \cdot s_n = 0$ and $c \cdot L = 0$. The limit becomes $\lim_{n \to \infty} 0 = 0$, which is trivially true. For any $\epsilon > 0$, we can choose any N, because $|0 - 0| < \epsilon$ is always true.
4. Case 2: c ≠ 0: If c is not equal to 0, we can proceed as follows. We want to get $|c \cdot s_n - c \cdot L|$ to be less than $\epsilon$. Let's manipulate this expression using the properties of absolute values: $|c \cdot s_n - c \cdot L| = |c(s_n - L)| = |c| \cdot |s_n - L|$.
5. Finding the Right N: Now, we want $|c| \cdot |s_n - L| < \epsilon$. Since we know that $|s_n - L|$ can be made arbitrarily small (because $\lim_{n \to \infty} s_n = L$), we can choose $\frac{\epsilon}{|c|}$ (since c is not 0, we can divide by |c|). So, for the given $\epsilon > 0$, there exists an $N_1$ such that if $n > N_1$, then $|s_n - L| < \frac{\epsilon}{|c|}$. This follows directly from the definition of the limit of $s_n$.
6. Putting it Together: Therefore, if $n > N_1$, then $|c \cdot s_n - c \cdot L| = |c| \cdot |s_n - L| < |c| \cdot \frac{\epsilon}{|c|} = \epsilon$. This proves that for any $\epsilon > 0$, there exists an $N_1$ such that if $n > N_1$, then $|c \cdot s_n - c \cdot L| < \epsilon$. This, by the definition of a limit, means that $\lim_{n \to \infty} (c \cdot s_n) = c \cdot L$.

Conclusion: The Power of the Limit Rule

And there you have it! We've successfully proven that $\lim_{n \to \infty} (c \cdot s_n) = c \cdot L$. This proof is a testament to the power of the formal definition of a limit and the properties of absolute values. It allows us to work with limits in a more predictable and manageable way. This seemingly simple result is incredibly useful in so many areas of math. This is an essential rule that you'll use over and over again as you keep going further with real analysis. I hope this explanation helped, guys. Keep practicing, and you'll master these concepts in no time. Good luck!