Suite Convergente : Limite De $(u_n = \frac{2n^2 + 1}{n^2 + 3n})$

Hey guys! Today, we're diving deep into the fascinating world of sequences and their limits. We've got a cool problem on our hands: to show that a specific sequence, $u_n = \frac{2n^2 + 1}{n^2 + 3n}$, converges, and then to figure out exactly what that limit is. This is a classic exercise that really helps solidify your understanding of how sequences behave as they go on and on, towards infinity. So, grab your notebooks, and let's break this down step by step. We'll make sure you understand every bit of it, so by the end, you'll be a pro at tackling similar problems!

Partie 1 : Démontrer la Convergence de la Suite $(u_n)$

Alright, first things first, we need to prove that our sequence, $u_n = \frac{2n^2 + 1}{n^2 + 3n}$, actually converges. What does it mean for a sequence to converge, you ask? Simply put, it means that as 'n' (that's our index, the number of the term in the sequence) gets bigger and bigger, the terms of the sequence get closer and closer to a specific, single number. Think of it like walking towards a destination; convergence means you're definitely getting there, not just wandering off into the wilderness. There are a few ways to prove convergence, but for rational functions like ours (where we have polynomials in the numerator and denominator), a common and super effective method is to analyze the behavior of the terms as $n$ approaches infinity. We can do this by dividing both the numerator and the denominator by the highest power of $n$ present in the denominator. In our case, the highest power of $n$ in the denominator ($n^2 + 3n$) is $n^2$. So, let's do that algebraic magic!

We rewrite $u_n$ as follows:

$$u_n = \frac{2n^2 + 1}{n^2 + 3n} = \frac{\frac{2n^2}{n^2} + \frac{1}{n^2}}{\frac{n^2}{n^2} + \frac{3n}{n^2}}$$

Simplifying this gives us:

$$u_n = \frac{2 + \frac{1}{n^2}}{1 + \frac{3}{n}}$$

Now, let's think about what happens as $n$ gets really large. We're talking about $n \to \infty$. As $n$ grows, terms like $\frac{1}{n^2}$ and $\frac{3}{n}$ become incredibly small. Imagine $n$ is a million; $\frac{1}{n^2}$ is one in a trillion, and $\frac{3}{n}$ is three in a million. These fractions practically vanish, approaching zero. This is a fundamental concept when dealing with limits of sequences involving rational functions. So, as $n \to \infty$, we have $\frac{1}{n^2} \to 0$ and $\frac{3}{n} \to 0$.

Plugging these vanishing terms back into our rewritten expression for $u_n$, we get:

$$\lim_{n\to\infty} u_n = \frac{2 + 0}{1 + 0} = \frac{2}{1} = 2$$

Since we found a specific finite value (which is 2) that the terms of the sequence approach as $n$ goes to infinity, this mathematically proves that the sequence $(u_n)$ converges. We've successfully shown the first part of our mission, guys! The sequence doesn't just wander; it heads straight for the number 2. This algebraic manipulation is key – it isolates the terms that grow with $n$ from those that become negligible. It's like filtering out the noise to see the main signal. The ability to identify and manipulate these terms is crucial for mastering limit problems. Remember this technique; it's a lifesaver for many sequence and series problems!

Partie 2 : Déterminer la Limite de la Suite $(u_n)$

Fantastic! We've already done the heavy lifting in the previous section, but let's formally state the second part of our exercise: determining the precise limit of the sequence $u_n = \frac{2n^2 + 1}{n^2 + 3n}$. As we saw during our convergence proof, the core of finding the limit for rational functions lies in examining the dominant terms – the ones with the highest powers of $n$ – and understanding how they behave as $n$ tends towards infinity. We effectively simplified the problem by dividing every term in the numerator and denominator by $n^2$, the highest power of $n$ in the denominator.

Let's recap our rewritten sequence:

$$u_n = \frac{2 + \frac{1}{n^2}}{1 + \frac{3}{n}}$$

Now, we apply the limit as $n \to \infty$. The properties of limits tell us that the limit of a quotient is the quotient of the limits (provided the denominator's limit is not zero), and the limit of a sum is the sum of the limits. So, we can break this down:

1. Limit of the numerator: $\lim_{n\to\infty} (2 + \frac{1}{n^2})$. We know that $\lim_{n\to\infty} 2 = 2$ (the limit of a constant is the constant itself) and $\lim_{n\to\infty} \frac{1}{n^2} = 0$ (as $n$ becomes infinitely large, $\frac{1}{n^2}$ becomes infinitesimally small, approaching zero). Therefore, $\lim_{n\to\infty} (2 + \frac{1}{n^2}) = 2 + 0 = 2$.

2. Limit of the denominator: $\lim_{n\to\infty} (1 + \frac{3}{n})$. Similarly, $\lim_{n\to\infty} 1 = 1$ and $\lim_{n\to\infty} \frac{3}{n} = 0$ (for the same reason as above, $\frac{3}{n}$ approaches zero as $n$ goes to infinity). Thus, $\lim_{n\to\infty} (1 + \frac{3}{n}) = 1 + 0 = 1$.

3. Limit of the quotient: Now we combine these results. Since the limit of the denominator is 1 (which is not zero), we can proceed:

$$\lim_{n\to\infty} u_n = \frac{\lim_{n\to\infty} (2 + \frac{1}{n^2})}{\lim_{n\to\infty} (1 + \frac{3}{n})} = \frac{2}{1} = 2$$

So, there you have it! The limit of the sequence $u_n = \frac{2n^2 + 1}{n^2 + 3n}$ as $n$ approaches infinity is 2. This means that as we calculate more and more terms of this sequence (like $u_{100}$, $u_{1000}$, $u_{1000000}$), the values will get extremely close to 2. We've successfully completed both parts of the exercise: we've proven convergence and pinpointed the exact limit. This technique of dividing by the highest power of $n$ is a gold standard for finding limits of rational functions. It works because it effectively compares the 'growth rates' of the polynomials in the numerator and denominator. When the degree of the numerator is the same as the degree of the denominator, as it is here (both are 2), the limit is simply the ratio of the leading coefficients (which are 2 and 1, respectively). Pretty neat, huh?

Intuition Behind the Limit

Let's get a bit more intuitive about why this limit is 2. When $n$ gets really, really massive, the '+1' in the numerator ($2n^2 + 1$) becomes almost insignificant compared to the $2n^2$ term. Similarly, the '+3n' in the denominator ($n^2 + 3n$) is dwarfed by the $n^2$ term. So, for huge values of $n$, the fraction $u_n = \frac{2n^2 + 1}{n^2 + 3n}$ behaves very much like $\frac{2n^2}{n^2}$. And guess what $\frac{2n^2}{n^2}$ simplifies to? Yep, just 2! This intuitive shortcut confirms our rigorous mathematical result. It’s always good to have both the formal proof and a gut feeling about why the answer makes sense. The dominant terms dictate the long-term behavior of the sequence. In this case, $n^2$ is the dominant term in both the numerator and the denominator, and their coefficients are 2 and 1, leading directly to the limit of 2. This understanding helps you predict limits even before doing detailed calculations.

Alternative Methods (for advanced learners!)

While dividing by the highest power of $n$ is the most straightforward method here, it's worth knowing that other techniques exist. For instance, if you're familiar with L'Hôpital's Rule (though typically used for functions of a real variable, it can be adapted for sequences if you consider the corresponding function), you could apply it here. If we let $f(x) = \frac{2x^2 + 1}{x^2 + 3x}$, as $x \to \infty$, we get the indeterminate form $\frac{\infty}{\infty}$. Applying L'Hôpital's Rule once gives $\lim_{x\to\infty} \frac{4x}{2x + 3}$. This is still $\frac{\infty}{\infty}$, so we apply it again: $\lim_{x\to\infty} \frac{4}{2} = 2$. This confirms our result! However, for sequences, the algebraic manipulation method is generally preferred as it doesn't require calculus concepts like derivatives and is often more direct.

Another perspective comes from understanding the degrees of the polynomials. When the degree of the numerator is equal to the degree of the denominator in a rational function, the limit at infinity is always the ratio of the leading coefficients. Here, the degree of $2n^2 + 1$ is 2, and the degree of $n^2 + 3n$ is also 2. The leading coefficient of the numerator is 2, and the leading coefficient of the denominator is 1. Thus, the limit is $\frac{2}{1} = 2$. This is a powerful rule of thumb that can save you a lot of time on standardized tests or quick checks.

Conclusion

So, guys, we've successfully tackled the problem! We've rigorously demonstrated that the sequence $u_n = \frac{2n^2 + 1}{n^2 + 3n}$ converges, and we've precisely determined its limit to be 2. The key techniques involved rewriting the expression by dividing by the highest power of $n$ in the denominator and then analyzing the behavior of the remaining terms as $n$ approaches infinity. Remember this approach – it's your go-to strategy for many limit problems involving rational sequences. Keep practicing, and you'll become a limit-finding ninja in no time! Until next time, happy studying!