Determining Continuity Of A Piecewise Function G(x) At X = 0

In this article, we delve into the intricacies of determining the continuity of a piecewise function at a specific point. Our focus will be on the 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. We aim to rigorously investigate the conditions under which g(x) is continuous at x = 0. This exploration involves understanding the concept of continuity, evaluating limits, and applying L'Hôpital's Rule to handle indeterminate forms.

Understanding Continuity

Before diving into the specifics of our function, let's solidify our understanding of continuity. A function f(x) is said to be continuous at a point x = c if the following three conditions are met:

1. f(c) is defined (i.e., the function exists at the point).
2. The limit of f(x) as x approaches c exists (i.e., the left-hand limit equals the right-hand limit).
3. The limit of f(x) as x approaches c is equal to f(c).

In simpler terms, a function is continuous at a point if there are no breaks, jumps, or holes at that point. The graph of the function can be drawn without lifting your pen.

For our piecewise function g(x) to be continuous at x = 0, we need to ensure that 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) both exist, are equal to each other, and that a value can be assigned at x=0 that matches this limit. Since the function is not defined at x=0, continuity at x=0 hinges entirely on the limits existing and being equal.

Evaluating the Left-Hand Limit

To determine the left-hand limit, we consider the part of the function defined for x < 0:

$$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} \frac{1 - a^x + x a^x \ln a}{a^x x^2}$$

Direct substitution yields an indeterminate form of type 0/0. This is a classic scenario where L'Hôpital's Rule comes to our rescue. L'Hôpital's Rule states that if the limit of f(x)/g(x) as x approaches c results in an indeterminate form of 0/0 or ∞/∞, then:

$$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$

provided the limit on the right-hand side exists. We can apply this rule repeatedly until we obtain a determinate form.

Let's apply L'Hôpital's Rule to our left-hand limit. First, we need to find the derivatives of the numerator and the denominator:

• Numerator: f(x) = 1 - a^x + x a^x ln a
  • f'(x) = -a^x ln a + a^x ln a + x a^x (ln a)^2 = x a^x (ln a)^2
• Denominator: g(x) = a^x x^2
  • g'(x) = a^x ln a * x^2 + 2x a^x = a^x(x2 ln a + 2x)

Applying L'Hôpital's Rule once, we get:

$$\lim_{x \to 0^-} \frac{1 - a^x + x a^x \ln a}{a^x x^2} = \lim_{x \to 0^-} \frac{x a^x (\ln a)^2}{a^x(x^2 \ln a + 2x)} = \lim_{x \to 0^-} \frac{x (\ln a)^2}{x^2 \ln a + 2x} = \lim_{x \to 0^-} \frac{(\ln a)^2}{x \ln a + 2}$$

Now, direct substitution gives us:

$$\frac{(\ln a)^2}{0 + 2} = \frac{(\ln a)^2}{2}$$

Thus, the left-hand limit is (ln a)^2 / 2.

Evaluating the Right-Hand Limit

Next, we evaluate the right-hand limit, considering the part of the function defined for x > 0:

$$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} \frac{2^x a^x - x \ln 2 - x \ln a - 1}{x^2}$$

Again, direct substitution leads to the indeterminate form 0/0. We will apply L'Hôpital's Rule. Let's find the derivatives of the numerator and denominator:

• Numerator: f(x) = 2^x a^x - x ln 2 - x ln a - 1
  • f'(x) = 2^x a^x ln 2 + 2^x a^x ln a - ln 2 - ln a = 2^x a^x (ln 2 + ln a) - (ln 2 + ln a)
• Denominator: g(x) = x^2
  • g'(x) = 2x

Applying L'Hôpital's Rule once:

$$\lim_{x \to 0^+} \frac{2^x a^x - x \ln 2 - x \ln a - 1}{x^2} = \lim_{x \to 0^+} \frac{2^x a^x (\ln 2 + \ln a) - (\ln 2 + \ln a)}{2x}$$

Direct substitution still results in 0/0, so we apply L'Hôpital's Rule a second time:

• Numerator: f'(x) = 2^x a^x (ln 2 + ln a) - (ln 2 + ln a)
  • f''(x) = 2^x a^x (ln 2 + ln a)^2
• Denominator: g'(x) = 2x
  • g''(x) = 2

Applying L'Hôpital's Rule again:

$$\lim_{x \to 0^+} \frac{2^x a^x (\ln 2 + \ln a) - (\ln 2 + \ln a)}{2x} = \lim_{x \to 0^+} \frac{2^x a^x (\ln 2 + \ln a)^2}{2}$$

Now, direct substitution gives us:

$$\frac{1 * 1 * (\ln 2 + \ln a)^2}{2} = \frac{(\ln 2 + \ln a)^2}{2}$$

Thus, the right-hand limit is (ln 2 + ln a)^2 / 2.

Condition for Continuity

For g(x) to be continuous at x = 0, the left-hand limit must equal the right-hand limit:

$$\frac{(\ln a)^2}{2} = \frac{(\ln 2 + \ln a)^2}{2}$$

Multiplying both sides by 2, we get:

$$(\ln a)^2 = (\ln 2 + \ln a)^2$$

Expanding the right side:

$$(\ln a)^2 = (\ln 2)^2 + 2 \ln 2 \ln a + (\ln a)^2$$

Subtracting (ln a)^2 from both sides:

$$0 = (\ln 2)^2 + 2 \ln 2 \ln a$$

$$0 = \ln 2 (\ln 2 + 2 \ln a)$$

Since ln 2 is not zero, we must have:

$$\ln 2 + 2 \ln a = 0$$

$$2 \ln a = -\ln 2$$

$$\ln a^2 = \ln 2^{-1}$$

$$a^2 = \frac{1}{2}$$

Since a > 0:

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

Conclusion

The function g(x) is continuous at x = 0 if and only if a = √2 / 2. This result was obtained by carefully evaluating the left-hand and right-hand limits using L'Hôpital's Rule and then equating them to satisfy the condition for continuity. This analysis demonstrates the importance of understanding limits and applying appropriate techniques to determine the behavior of functions at specific points.

This detailed examination provides a comprehensive understanding of how to approach continuity problems for piecewise functions, emphasizing the crucial role of limits and L'Hôpital's Rule in resolving indeterminate forms.