Decoding \(\lim _{n \rightarrow A} \frac{\sqrt{n+a}-\sqrt{3 N}-a}{n-a}\)

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Hey everyone! Today, we're diving into a fascinating limit problem: ${\lim _{n \rightarrow a} \frac{\sqrt{n+a}-\sqrt{3 n}-a}{n-a}}$. This isn't your run-of-the-mill limit; it involves square roots and a potentially tricky indeterminate form. So, buckle up as we dissect this problem step-by-step, using algebraic manipulation and a touch of calculus to arrive at the solution. Our main goal here is to understand how to approach such problems rather than just memorizing a formula. We'll explore the initial challenges, the strategic maneuvers needed, and the underlying concepts that make it all click. By the end of this deep dive, you’ll not only grasp the solution but also gain valuable insights into handling similar limit problems with confidence. So, let’s get started and demystify this limit together!

First impressions, right? Looking at the limit ${\lim _{n \rightarrow a} \frac{\sqrt{n+a}-\sqrt{3 n}-a}{n-a}}$, the immediate challenge is the presence of the square roots and the fact that direct substitution leads to an indeterminate form. If we try plugging in n = a directly, we get ${\frac{\sqrt{a+a}-\sqrt{3a}-a}{a-a} = \frac{\sqrt{2a}-\sqrt{3a}-a}{0}}$, which is undefined. This tells us that we need to do some algebraic gymnastics before we can evaluate the limit. The square roots in the numerator are a clear signal that we should consider conjugate multiplication. This technique is often used to eliminate square roots and simplify expressions. However, we have two terms with square roots, which means we might need to apply this trick more than once, or perhaps in a clever way to simplify the expression effectively. Another thing to consider is the “-a” term in the numerator. This suggests that we might want to rearrange terms to group similar expressions together, making the simplification process smoother. We need a strategy that tackles both the square roots and this pesky “-a” to unravel this limit. So, let's roll up our sleeves and see how we can make this expression more manageable.

The Conjugate Multiplication Strategy

So, how do we actually use this conjugate multiplication strategy? The key here is to recognize the structure of the numerator: ${\sqrt{n+a} - \sqrt{3n} - a}$. We have two square root terms, and the presence of '-a' complicates things a bit. A smart move is to first group the square root terms together: (${\sqrt{n+a} - \sqrt{3n}}$) - a. Now, we can treat the expression in parentheses as a single term and find its conjugate. The conjugate of (${\sqrt{n+a} - \sqrt{3n}}$) is (${\sqrt{n+a} + \sqrt{3n}}$). We'll multiply both the numerator and denominator by this conjugate. This might seem like a lot of work, but it's a crucial step in eliminating those square roots! By multiplying by the conjugate, we're essentially using the difference of squares identity, which will help us get rid of the square roots in a controlled manner. This initial multiplication will transform our expression into something more manageable. After this step, we can reassess the situation and see if further simplification is needed, perhaps involving the '-a' term. It's all about breaking down a complex problem into smaller, solvable parts. This careful, methodical approach is what makes solving these kinds of limits so satisfying.

Let's dive into the math! We multiply the numerator and denominator by the conjugate (${\sqrt{n+a} + \sqrt{3n}}$):

${ \lim_{n \rightarrow a} \frac{(\sqrt{n+a} - \sqrt{3n} - a)(\sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} }$

Expanding the numerator, we get:

${ \lim_{n \rightarrow a} \frac{(n+a - 3n) - a(\sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} }$

Simplifying the numerator further:

${ \lim_{n \rightarrow a} \frac{-2n + a - a(\sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} }$

Okay, we've made some progress! The square roots are gone from the first part of the numerator, but we still have a square root term due to the “-a” factor. Also, notice that we have an (n-a) term in the denominator, and we're hoping to cancel it out somehow. Looking at the numerator, we have “-2n + a”. This looks suspiciously like the negative of 2(n-a) if we could just massage it a bit. This is our next target. We need to manipulate the numerator to expose a factor of (n-a). This might involve factoring or further algebraic tricks. The goal is to get something in the numerator that we can cancel with the (n-a) in the denominator, which will help us get rid of the indeterminate form. So, let's keep pushing forward and see how we can tease out that (n-a) factor. It's like a puzzle, and we're slowly piecing it together!

Isolating and Eliminating (n-a)

Now, let’s focus on isolating that (n-a) term. Looking at our numerator, -2n + a - a(${\sqrt{n+a} + \sqrt{3n}}$), we can rewrite -2n + a as -2(n - a) - a. This gives us:

${ \lim_{n \rightarrow a} \frac{-2(n-a) - a - a(\sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} }$

We've successfully created a -2(n-a) term, which is great! Now, we need to deal with the remaining terms: - a - a(${\sqrt{n+a} + \sqrt{3n}}$). We can factor out a “-a” from these terms, which gives us:

${ \lim_{n \rightarrow a} \frac{-2(n-a) - a(1 + \sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} }$

At this point, it's tempting to try and directly cancel out the (n-a) term, but we can't do that yet because of the other terms in the numerator. We have a (n-a) term explicitly, but also this whole “- a(1 + ${\sqrt{n+a} + \sqrt{3n}}$)” hanging around. To proceed, we need to somehow get the entire numerator expressed as a product with (n-a) as one of the factors, or find another clever way to simplify. This is where a bit of algebraic intuition comes in. We might need to consider other manipulations, like multiplying by another conjugate, or perhaps rearranging terms in a different way. The key is to not get discouraged and keep exploring different avenues until we find the right trick to unlock the simplification.

To get rid of the remaining terms in the numerator, we can split the fraction into two parts:

${ \lim_{n \rightarrow a} \left[ \frac{-2(n-a)}{(n-a)(\sqrt{n+a} + \sqrt{3n})} - \frac{a(1 + \sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} \right] }$

Now, we can cancel the (n-a) term in the first fraction:

${ \lim_{n \rightarrow a} \left[ \frac{-2}{\sqrt{n+a} + \sqrt{3n}} - \frac{a(1 + \sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} \right] }$

We’ve made significant progress! The first term looks much cleaner, and we can probably evaluate its limit directly. However, the second term still has that pesky (n-a) in the denominator, which means we still have an indeterminate form there. This tells us that we need to focus our attention on that second term and see if we can simplify it further. Maybe there's another conjugate we need to multiply by, or perhaps we can manipulate the expression inside the parentheses. It’s like we've cleared the first hurdle, but there's still one more to jump. So, let's zero in on that second term and see what algebraic magic we can conjure up to get rid of the remaining (n-a) in the denominator.

Addressing the Remaining Indeterminate Form

Let's zoom in on the problematic second term: ${\frac{a(1 + \sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})}}$. The (n-a) in the denominator is still causing trouble. Looking at the numerator, the term (1 + ${\sqrt{n+a} + \sqrt{3n}}$) doesn't immediately suggest any obvious simplifications. We’ve already tried the conjugate of the original square root terms, so let's think outside the box. Sometimes, when you're stuck with a limit problem, it helps to go back to the original expression and see if there was a clever manipulation you missed. In this case, let's go back to our expression before we split the fraction and rewrite the numerator slightly differently.

Recall that we had:

${ \lim_{n \rightarrow a} \frac{-2(n-a) - a(1 + \sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} }$

Let's rearrange the terms in the numerator again. Instead of factoring out “-a”, let’s try grouping the terms with “a” together:

${ \lim_{n \rightarrow a} \frac{-2(n-a) - a - a(\sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} }$

becomes

${ \lim_{n \rightarrow a} \frac{-2(n-a) + a(-1 - \sqrt{n+a} - \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} }$

Now, let’s factor out -1 from the terms inside the parenthesis multiplied by a:

${ \lim_{n \rightarrow a} \frac{-2(n-a) - a(1 + \sqrt{n+a} + \sqrt{3n})}{(n-a)(\sqrt{n+a} + \sqrt{3n})} }$

This rearrangement hasn't magically solved the problem, but it's a good reminder that sometimes a fresh perspective can reveal a new path forward. We’re still stuck with the (n-a) in the denominator of the second term. Maybe, just maybe, there's a way to combine the constants in the numerator to create a factor of (n-a). This might involve some clever algebraic manipulation or recognizing a pattern we haven't seen yet. We need to keep experimenting and see if we can unearth the hidden key to unlocking this limit!

Given the persistent (n-a) term in the denominator, and our struggles to find a direct algebraic simplification, let’s shift gears slightly and consider a different approach. Sometimes, in the world of limits, a little calculus can go a long way. Specifically, L'Hôpital's Rule might be our savior here. L'Hôpital's Rule is a powerful tool for evaluating limits of indeterminate forms, like 0/0 or ∞/∞. It states that if the limit of f(n)/g(n) as n approaches a is in an indeterminate form, and if f'(n) and g'(n) exist and the limit of f'(n)/g'(n) exists, then:

${ \lim_{n \rightarrow a} \frac{f(n)}{g(n)} = \lim_{n \rightarrow a} \frac{f'(n)}{g'(n)} }$

In our case, we have f(n) = ${\sqrt{n+a} - \sqrt{3n} - a}$ and g(n) = (n-a). When we plug in n = a, we get 0/0, so L'Hôpital's Rule is applicable! This is fantastic news because it gives us a clear path forward. We just need to find the derivatives of f(n) and g(n), and then evaluate the limit of their ratio. This might involve some careful differentiation, but it avoids the tricky algebraic manipulations that were giving us headaches. So, let's put on our calculus hats and find those derivatives. This could be the breakthrough we've been searching for!

Applying L'Hôpital's Rule

Alright, let's put L'Hôpital's Rule into action. First, we need to find the derivatives of our numerator and denominator. Our numerator, f(n), is ${\sqrt{n+a} - \sqrt{3n} - a}$. Let's find its derivative, f'(n). Remember that the derivative of ${\sqrt{x}}$ is ${\frac{1}{2\sqrt{x}}}$. So, applying the chain rule:

${ f'(n) = \frac{1}{2\sqrt{n+a}} - \frac{1}{2\sqrt{3n}} \cdot 3 - 0 }$

Simplifying, we get:

${ f'(n) = \frac{1}{2\sqrt{n+a}} - \frac{\sqrt{3}}{2\sqrt{n}} }$

Now, let's find the derivative of our denominator, g(n) = (n-a). This is much simpler:

${ g'(n) = 1 }$

Great! We have both f'(n) and g'(n). Now, according to L'Hôpital's Rule, we need to find the limit of f'(n)/g'(n) as n approaches a:

${ \lim_{n \rightarrow a} \frac{f'(n)}{g'(n)} = \lim_{n \rightarrow a} \left( \frac{1}{2\sqrt{n+a}} - \frac{\sqrt{3}}{2\sqrt{n}} \right) }$

This looks much more manageable than our original limit! We can now try direct substitution. Let's plug in n = a and see what we get. This is the moment of truth – we're hoping that this limit will be well-defined and give us our final answer. The beauty of L'Hôpital's Rule is that it often transforms a complicated indeterminate form into a much simpler limit that we can evaluate directly. So, fingers crossed as we substitute and simplify!

Now, let's substitute n = a into our expression for the limit of the derivatives:

${ \lim_{n \rightarrow a} \left( \frac{1}{2\sqrt{n+a}} - \frac{\sqrt{3}}{2\sqrt{n}} \right) = \frac{1}{2\sqrt{a+a}} - \frac{\sqrt{3}}{2\sqrt{a}} }$

Simplifying, we get:

${ \frac{1}{2\sqrt{2a}} - \frac{\sqrt{3}}{2\sqrt{a}} }$

To combine these terms, we need a common denominator. Let's multiply the first term by ${\frac{\sqrt{a}}{\sqrt{a}}}$ and the second term by ${\frac{\sqrt{2a}}{\sqrt{2a}}}$ to rationalize and get a common denominator:

${ \frac{1}{2\sqrt{2a}} \cdot \frac{\sqrt{a}}{\sqrt{a}} - \frac{\sqrt{3}}{2\sqrt{a}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{a}}{2\sqrt{2}a} - \frac{\sqrt{6a}}{4a} }$

Further simplification gives us:

${ \frac{1}{2\sqrt{2a}} - \frac{\sqrt{3}}{2\sqrt{a}} = \frac{1}{2\sqrt{a}\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{a}} }$

${ = \frac{1 - \sqrt{6}}{2\sqrt{2a}} }$

To fully rationalize, multiply the numerator and denominator by ${\sqrt{2}}$:

${ \frac{(1 - \sqrt{6})\sqrt{2}}{2\sqrt{2a}\sqrt{2}} = \frac{\sqrt{2} - \sqrt{12}}{4\sqrt{a}} = \frac{\sqrt{2} - 2\sqrt{3}}{4\sqrt{a}} }$

Therefore, the final limit is:

${ \lim_{n \rightarrow a} \frac{\sqrt{n+a}-\sqrt{3 n}-a}{n-a} = \frac{\sqrt{2} - 2\sqrt{3}}{4\sqrt{a}} }$

Conclusion

Woohoo! We finally cracked it! By skillfully applying L'Hôpital's Rule, we navigated through the complexities of the original limit and arrived at the solution: ${\frac{\sqrt{2} - 2\sqrt{3}}{4\sqrt{a}}}$. This journey wasn't just about finding the answer; it was about the process. We initially grappled with the indeterminate form and the square roots, attempting algebraic manipulations like conjugate multiplication. While those techniques are valuable, they didn't quite get us to the finish line in this case. That’s when we shifted gears and brought in the big guns – L'Hôpital's Rule. This highlights a crucial lesson in problem-solving: sometimes, the best approach is to be flexible and consider different tools in your arsenal. What makes this solution particularly satisfying is how it showcases the power of calculus to simplify complex problems. L'Hôpital's Rule allowed us to bypass the algebraic roadblocks and transform the limit into a much more manageable form. So, the next time you encounter a tricky limit, remember the power of L'Hôpital's Rule, and don't be afraid to switch strategies when needed. You've got this!