Determining Limits The Case Of Lim (x, Y)→(0,0) 2x³y / (x⁶ + 3y²)

Jul 30, 2025 by Chloe Fitzgerald 66 views

$Determine if Limits Exist and Find Their Values: A Deep Dive into $\lim _{(x, y) \rightarrow(0,0)} \frac{2 x^3 y}{x^6+3 y^2}$ ## Understanding the Limit Problem Hey guys! Let's tackle a classic multivariable calculus problem today – determining whether a limit exists and, if it does, finding its value. We're specifically looking at this limit: $\lim _{(x, y) \rightarrow(0,0)} \frac{2 x^3 y}{x^6+3 y^2}$ This isn't your typical single-variable limit, so we need to get creative. The challenge here is that as (x, y) approaches (0, 0), there are infinitely many paths we could take. For the limit to exist, we need to get the same value no matter which path we choose. If we find even *one* pair of paths that give us different limits, we know the overall limit doesn't exist. So, what's our strategy? We'll explore different paths towards (0, 0) and see if the function converges to the same value. If it does, that's a good sign, but not a guarantee. If it doesn't, we know the limit doesn't exist. Let's start with some simple paths, like approaching along the x-axis and y-axis. We'll also try approaching along lines like y = mx. If those give us the same limit, we might need to try something more sophisticated, like a parabolic path. The goal is to find a path that exposes any potential inconsistencies. This problem really highlights the difference between single-variable and multivariable calculus. In single-variable calculus, there are only two directions to approach a point (left and right). But in multivariable calculus, the possibilities are endless! This makes limit evaluation much more nuanced and interesting. We have to be thorough and methodical to make sure we've explored enough paths to make a confident conclusion. So, stick with me as we dive in and explore this limit from every angle! We'll break it down step by step and make sure you understand the techniques involved. By the end, you'll be a pro at tackling these tricky multivariable limits. ## Exploring Paths to (0,0) Guys, let's get our hands dirty and start exploring some paths! Remember, our goal is to see if the function $ rac{2 x^3 y}{x^6+3 y^2}$ approaches the same value as (x, y) approaches (0, 0) along different routes. If we find a discrepancy, the limit doesn't exist. **Path 1: Along the x-axis (y = 0)** This is a classic starting point. If we set y = 0, our function becomes: $\frac{2 x^3 (0)}{x^6+3 (0)^2} = \frac{0}{x^6} = 0$ As x approaches 0 along the x-axis, the function is always 0. So, the limit along this path is 0. Okay, that's one data point. Now let's try another path. **Path 2: Along the y-axis (x = 0)** Similarly, let's see what happens when we approach along the y-axis, setting x = 0: $\frac{2 (0)^3 y}{(0)^6+3 y^2} = \frac{0}{3 y^2} = 0$ Again, the function simplifies to 0 as y approaches 0 along the y-axis. So, the limit along this path is also 0. So far, so good! Both paths give us a limit of 0. But remember, this doesn't guarantee the limit exists. We need to explore more paths. **Path 3: Along the line y = mx** Let's try a more general path – a straight line passing through the origin, y = mx, where m is a constant. Substituting y = mx into our function, we get: $\frac{2 x^3 (mx)}{x^6+3 (mx)^2} = \frac{2 m x^4}{x^6+3 m^2 x^2}$ Now, we can simplify this by dividing both the numerator and denominator by x^2: $\frac{2 m x^2}{x^4+3 m^2}$ Now, as x approaches 0, the numerator approaches 0, and the denominator approaches 3m^2. Therefore, the limit along the line y = mx is: $\lim _{x \rightarrow 0} \frac{2 m x^2}{x^4+3 m^2} = \frac{0}{3 m^2} = 0$ (assuming m is not 0). If m = 0, we're back to the x-axis, which we already handled. Wow! Three paths, and they all give us a limit of 0. This is making us think the limit might actually exist. But we can't get complacent. We need to keep searching for a potential troublemaker path. Let's try something a little different... ## Unveiling the Truth: A Parabolic Path Now guys, we've explored some basic paths – the axes and lines – and they all pointed to a limit of 0. But remember, multivariable limits can be sneaky. Sometimes, the limit behaves nicely along straight lines but does something completely different along curves. So, let's crank up the difficulty and explore a parabolic path. **Path 4: Along the curve y = x^3** This is where things get interesting! Let's substitute y = x^3 into our function: $\frac{2 x^3 (x^3)}{x^6+3 (x^3)^2} = \frac{2 x^6}{x^6+3 x^6} = \frac{2 x^6}{4 x^6} = \frac{1}{2}$ Notice something amazing? The x^6 terms cancel out completely! This means that *along the path y = x^3, the function is always equal to 1/2*, regardless of how close we are to (0, 0). Therefore, the limit along this path is: $\lim _{x \rightarrow 0} \frac{2 x^3 (x^3)}{x^6+3 (x^3)^2} = \frac{1}{2}$ Hold on a second! We've got a problem. We found a path (y = x^3) where the limit is 1/2, but we previously found paths (x-axis, y-axis, and y = mx) where the limit was 0. This is a **huge red flag**. Remember, for a limit to exist, it *must* be the same along *every* possible path. Since we've found two paths that give us different limits, we can confidently conclude that the limit does **not** exist. This is the beauty (and sometimes the frustration) of multivariable limits. They force us to be rigorous and explore all possibilities. This parabolic path was the key to unlocking the truth. It showed us that the function's behavior is more complex than we initially thought. Let's summarize our findings and wrap things up. ## Conclusion: The Limit Does Not Exist Alright guys, let's recap what we've discovered. We started with the limit: $\lim _{(x, y) \rightarrow(0,0)} \frac{2 x^3 y}{x^6+3 y^2}$ We embarked on a journey to explore different paths to (0, 0) to see if the function converged to a single value. * We first approached along the x-axis (y = 0) and found a limit of 0. * Then, we approached along the y-axis (x = 0) and again found a limit of 0. * Next, we tackled the lines y = mx and consistently found a limit of 0. It seemed like the limit might exist and be equal to 0. But we knew we couldn't stop there. * Finally, we ventured onto the parabolic path y = x^3, and *bam!* We found a limit of 1/2. This was the critical moment. The fact that we found *two different limits* along different paths is the nail in the coffin. **Therefore, we can definitively conclude that the limit $\lim _{(x, y) \rightarrow(0,0)} \frac{2 x^3 y}{x^6+3 y^2}$ does not exist.** This problem is a fantastic illustration of why multivariable limits are so interesting and challenging. We can't just approach along a few simple paths and call it a day. We need to be persistent and creative in our exploration. The key takeaway here is that *if you find even one pair of paths that lead to different limits, the overall limit does not exist*. This is a fundamental principle in multivariable calculus, and mastering it will help you conquer even the trickiest limit problems. So, keep practicing, keep exploring, and never give up on the quest to understand these fascinating mathematical concepts!$