Continuity Analysis Of A Piecewise Function G(x) At X = 0
In the fascinating world of mathematical analysis, understanding the behavior of functions, especially around specific points, is crucial. A key concept in this realm is continuity. A function is said to be continuous at a point if its limit exists at that point, its value is defined at that point, and the limit is equal to the function's value. This article delves into a detailed analysis of the continuity of a piecewise-defined function, focusing on the intricate steps required to ensure such continuity. We will explore the given function,
g(x) = \begin{cases} \frac{1 - a^x + x a^x \ln a}{a^x x^2} & \text{for } x < 0 \\ \frac{2^x a^x - x \ln 2 - x \ln a - 1}{x^2} & \text{for } x > 0 \end{cases}
where a > 0, at the point x = 0. This exploration involves evaluating limits, applying L'Hôpital's Rule, and ensuring that the left-hand limit, right-hand limit, and the function's value at x = 0 all coincide. The journey through this problem will highlight the elegance and power of calculus in dissecting and understanding the properties of functions. In our exploration, we'll meticulously examine each part of the piecewise function, paying close attention to how the parameter a influences the function's behavior near x = 0. Our primary objective is to determine the conditions under which g(x) maintains continuity at this critical juncture. This process not only provides insight into the specific function at hand but also offers a broader understanding of continuity, limits, and the application of advanced calculus techniques.
We are given the piecewise function:
g(x) = \begin{cases} \frac{1 - a^x + x a^x \ln a}{a^x x^2} & \text{for } x < 0 \\ \frac{2^x a^x - x \ln 2 - x \ln a - 1}{x^2} & \text{for } x > 0 \end{cases}
where a > 0. Our goal is to determine the conditions for g(x) to be continuous at x = 0. This involves analyzing the left-hand limit (as x approaches 0 from the negative side) and the right-hand limit (as x approaches 0 from the positive side). For g(x) to be continuous at x = 0, these limits must exist, be equal, and the function must be defined at x = 0 with its value matching the limits. This problem is an excellent example of how continuity is ensured for a piecewise function, requiring a careful examination of limits and the application of relevant calculus principles. Understanding this problem not only enhances our grasp of continuity but also improves our ability to handle similar analytical challenges in calculus.
To investigate the left-hand limit of g(x) as x approaches 0, we focus on the part of the function defined for x < 0:
g(x) = \frac{1 - a^x + x a^x \ln a}{a^x x^2}
We need to evaluate:
\lim_{x \to 0^-} \frac{1 - a^x + x a^x \ln a}{a^x x^2}
Direct substitution of x = 0 yields an indeterminate form of type 0/0, which calls for the application of L'Hôpital's Rule. This rule allows us to differentiate the numerator and the denominator separately and then re-evaluate the limit. Let's first compute the derivatives of the numerator and the denominator.
The numerator is 1 - a^x + x a^x ln a, and its derivative with respect to x is:
-\frac{d}{dx}(a^x) + \frac{d}{dx}(x a^x \ln a) = -a^x \ln a + a^x \ln a + x a^x (\ln a)^2 = x a^x (\ln a)^2
The denominator is a^x x^2, and its derivative with respect to x is:
\frac{d}{dx}(a^x x^2) = a^x \ln a \cdot x^2 + 2x a^x = a^x x (x \ln a + 2)
Applying L'Hôpital's Rule once, we get:
\lim_{x \to 0^-} \frac{x a^x (\ln a)^2}{a^x x (x \ln a + 2)} = \lim_{x \to 0^-} \frac{(\ln a)^2}{x \ln a + 2}
Now, substituting x = 0 gives us:
\frac{(\ln a)^2}{2}
Thus, the left-hand limit exists and is equal to (ln a)^2 / 2. This result is a crucial step in ensuring the continuity of g(x) at x = 0. The existence of this limit means that as we approach x = 0 from the negative side, the function g(x) converges to this specific value. However, for g(x) to be continuous, the right-hand limit must also exist and be equal to this value. This sets the stage for our next investigation, where we analyze the behavior of g(x) as x approaches 0 from the positive side.
Next, we analyze the right-hand limit of g(x) as x approaches 0. This involves the part of the function defined for x > 0:
g(x) = \frac{2^x a^x - x \ln 2 - x \ln a - 1}{x^2}
We need to evaluate:
\lim_{x \to 0^+} \frac{2^x a^x - x \ln 2 - x \ln a - 1}{x^2}
Again, direct substitution of x = 0 yields an indeterminate form of type 0/0, indicating the necessity of using L'Hôpital's Rule. We differentiate the numerator and the denominator separately.
The numerator is 2^x a^x - x ln 2 - x ln a - 1, and its derivative with respect to x is:
\frac{d}{dx}(2^x a^x) - \frac{d}{dx}(x \ln 2) - \frac{d}{dx}(x \ln a) - \frac{d}{dx}(1) = 2^x a^x (\ln 2 + \ln a) - \ln 2 - \ln a
The denominator is x^2, and its derivative with respect to x is:
\frac{d}{dx}(x^2) = 2x
Applying L'Hôpital's Rule once, we get:
\lim_{x \to 0^+} \frac{2^x a^x (\ln 2 + \ln a) - \ln 2 - \ln a}{2x}
However, direct substitution still yields an indeterminate form of type 0/0. Therefore, we apply L'Hôpital's Rule again. Differentiating the numerator, we get:
\frac{d}{dx}(2^x a^x (\ln 2 + \ln a) - \ln 2 - \ln a) = 2^x a^x (\ln 2 + \ln a)^2
The derivative of the denominator is:
\frac{d}{dx}(2x) = 2
Applying L'Hôpital's Rule a second time, we have:
\lim_{x \to 0^+} \frac{2^x a^x (\ln 2 + \ln a)^2}{2}
Now, substituting x = 0 gives us:
\frac{(\ln 2 + \ln a)^2}{2}
Thus, the right-hand limit exists and is equal to (ln 2 + ln a)^2 / 2. This result is crucial for determining the continuity of g(x) at x = 0. The existence of this limit means that as we approach x = 0 from the positive side, the function g(x) converges to this specific value. For g(x) to be continuous at x = 0, this right-hand limit must be equal to the left-hand limit we computed earlier. This equality is the condition we will explore in the next section.
For the function g(x) to be continuous at x = 0, the left-hand limit and the right-hand limit must be equal. We have found that:
- The left-hand limit is
(\ln a)^2 / 2 - The right-hand limit is
(\ln 2 + \ln a)^2 / 2
Therefore, we must have:
\frac{(\ln a)^2}{2} = \frac{(\ln 2 + \ln a)^2}{2}
Multiplying both sides by 2, we get:
(\ln a)^2 = (\ln 2 + \ln a)^2
Expanding the right side, we have:
(\ln a)^2 = (\ln 2)^2 + 2 \ln 2 \ln a + (\ln a)^2
Subtracting (ln a)^2 from both sides, we get:
0 = (\ln 2)^2 + 2 \ln 2 \ln a
We can rewrite this as:
2 \ln 2 \ln a = -(\ln 2)^2
Since ln 2 is not zero, we can divide both sides by 2 ln 2:
\ln a = -\frac{\ln 2}{2} = \ln(2^{-1/2})
Therefore, we find that:
a = 2^{-1/2} = \frac{1}{\sqrt{2}}
This is the condition on a for which the function g(x) is continuous at x = 0. This value of a ensures that the left and right limits of the function at x = 0 are the same, which is a fundamental requirement for continuity. The process of arriving at this condition highlights the importance of applying L'Hôpital's Rule in indeterminate forms and the algebraic manipulation required to solve for specific conditions that ensure continuity.
In conclusion, we have meticulously analyzed the piecewise function g(x) to determine the condition for its continuity at x = 0. By independently evaluating the left-hand and right-hand limits using L'Hôpital's Rule and subsequently equating them, we found that the function g(x) is continuous at x = 0 if and only if a = 1/√2. This result underscores the significance of understanding limits and their role in establishing the continuity of functions, especially piecewise-defined ones.
The process of solving this problem involved several key steps: first, we identified the need to evaluate the left-hand and right-hand limits due to the piecewise nature of the function. Second, we applied L'Hôpital's Rule to resolve the indeterminate forms that arose when directly substituting x = 0 into the function. This step was crucial in obtaining manageable expressions for the limits. Third, we equated the left-hand and right-hand limits and solved the resulting equation for a, which provided the condition for continuity. This entire process demonstrates the rigorous approach required in mathematical analysis to ensure the continuity of functions.
The implications of this result extend beyond this specific problem. Understanding how to analyze the continuity of piecewise functions is essential in various fields, including engineering, physics, and computer science, where such functions are often used to model real-world phenomena. The techniques used here, such as L'Hôpital's Rule and limit evaluation, are fundamental tools in calculus and analysis, and mastering them is crucial for anyone working with continuous functions. This exploration not only provides a solution to the given problem but also reinforces the importance of these analytical skills in broader scientific and mathematical contexts.
