Mastering Limits How To Solve $\lim _{x \rightarrow A} \frac{\sqrt{x+a}-\sqrt{3 X-a}}{x-a}$

Hey guys! Today, we're diving deep into a fascinating limit problem that often pops up in calculus: $\lim _{x \rightarrow a} \frac{\sqrt{x+a}-\sqrt{3 x-a}}{x-a}$. This isn't your everyday limit; it requires a bit of algebraic finesse to crack. So, buckle up, and let's get started!

Understanding the Indeterminate Form

First things first, let's talk about what happens when we directly substitute x = a into the expression. If we do that, we get:

$\frac{\sqrt{a+a}-\sqrt{3a-a}}{a-a} = \frac{\sqrt{2a}-\sqrt{2a}}{0} = \frac{0}{0}$

Ah, the infamous 0/0 indeterminate form! This tells us that we can't just plug and chug; we need to do some algebraic manipulation to reveal the true nature of the limit. Indeterminate forms are common in calculus, and they signal that there's more to the limit than meets the eye. In this case, the 0/0 form suggests that there's a common factor in the numerator and denominator that we can cancel out, but we need to find it first.

The Conjugate Trick: Our Key to Success

Now, here's where the magic happens. When you see square roots in a limit problem, especially when dealing with indeterminate forms, the conjugate is your best friend. The conjugate is simply the expression you get by changing the sign between the terms in the numerator (or denominator, if the roots are there). In our case, the conjugate of $\sqrt{x+a} - \sqrt{3x-a}$ is $\sqrt{x+a} + \sqrt{3x-a}$.

So, what do we do with it? We multiply both the numerator and the denominator of our original expression by this conjugate. This might seem like we're just making things more complicated, but trust me, it's a brilliant move. By multiplying by the conjugate, we're essentially multiplying by 1 (since anything divided by itself is 1), so we're not changing the value of the expression. However, this clever manipulation will help us get rid of the square roots in the numerator.

Here's how it looks:

$\lim _{x \rightarrow a} \frac{\sqrt{x+a}-\sqrt{3 x-a}}{x-a} \cdot \frac{\sqrt{x+a}+\sqrt{3 x-a}}{\sqrt{x+a}+\sqrt{3 x-a}}$

When we multiply the numerators, we're using the difference of squares pattern: (A - B)(A + B) = A² - B². This is where the square roots start to disappear, which is exactly what we want!

Simplifying the Expression

Let's multiply out the numerator:

$(\sqrt{x+a} - \sqrt{3x-a})(\sqrt{x+a} + \sqrt{3x-a}) = (\sqrt{x+a})^2 - (\sqrt{3x-a})^2 = (x+a) - (3x-a)$

Simplifying further, we get:

$x + a - 3x + a = -2x + 2a$

Now, let's rewrite our limit with the simplified numerator:

$\lim _{x \rightarrow a} \frac{-2x + 2a}{(x-a)(\sqrt{x+a}+\sqrt{3 x-a})}$

Notice anything interesting? We can factor out a -2 from the numerator:

$\lim _{x \rightarrow a} \frac{-2(x - a)}{(x-a)(\sqrt{x+a}+\sqrt{3 x-a})}$

Boom! We've got a common factor of (x - a) in both the numerator and the denominator. This is the factor that was causing the 0/0 indeterminate form, and now we can finally get rid of it.

Canceling the Common Factor

We can now cancel out the (x - a) terms:

$\lim _{x \rightarrow a} \frac{-2}{\sqrt{x+a}+\sqrt{3 x-a}}$

This looks much better! We've eliminated the problematic factor, and now we can try substituting x = a again.

Evaluating the Limit

Let's substitute x = a into the simplified expression:

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

Simplifying, we get:

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

And there you have it! The limit is $\frac{-1}{\sqrt{2a}}$.

Wrapping Up the Solution

So, to recap, we started with the limit $\lim _{x \rightarrow a} \frac{\sqrt{x+a}-\sqrt{3 x-a}}{x-a}$, which gave us the indeterminate form 0/0. We then used the conjugate trick to rationalize the numerator, simplified the expression, canceled out the common factor, and finally evaluated the limit. The result is $\frac{-1}{\sqrt{2a}}$. This problem showcases a classic technique for dealing with limits involving square roots, and it's a valuable tool to have in your calculus arsenal.

Alright, let's broaden our horizons a bit and talk about limits in general. Understanding limits is crucial in calculus because it forms the foundation for concepts like derivatives and integrals. We've tackled a specific example, but there's a whole universe of limit problems out there, each with its unique challenges and solutions. So, let's equip ourselves with the knowledge and strategies to conquer them all!

The Essence of Limits: Approaching a Value

At its core, a limit describes the value that a function approaches as the input (usually x) gets closer and closer to a certain value. It's not necessarily about the value of the function at that point, but rather the trend as we get infinitely close. This distinction is vital, especially when dealing with functions that are undefined at a specific point, like in our previous example where substituting x = a directly resulted in 0/0. The limit is all about the journey, not necessarily the destination.

Think of it like this: Imagine you're walking towards a door. The limit is the position of the door, but you don't necessarily have to walk through the door to know where it is. You can get arbitrarily close to the door, and that's what the limit captures. This idea of