Analyzing Continuity of Functions

Let's analyze the continuity of the given function \( \rm f(x) = sin \left(\frac{1}{x^2}\right) \), for \( \rm x \ne 0 \). We need to evaluate the two statements provided.

Statement 1: Continuity at x = 0

The first statement claims that the function \( \rm f(x) \) is continuous at \( \rm x = 0 \) if \( \rm f(0) = 0 \). For a function to be continuous at a point \( \rm x = a \), three conditions must be met:

• \( \rm f(a) \) must be defined.
• The limit \( \rm \lim_{x \to a} f(x) \) must exist.
• The limit must be equal to the function value, i.e., \( \rm \lim_{x \to a} f(x) = f(a) \).

In this case, we are considering continuity at \( \rm x = 0 \). We are given that \( \rm f(0) = 0 \) is defined. However, we need to check if the limit \( \rm \lim_{x \to 0} f(x) = \lim_{x \to 0} sin \left(\frac{1}{x^2}\right) \) exists.

Let's consider the behavior of the argument of the sine function, \( \rm \frac{1}{x^2} \), as \( \rm x \to 0 \). As \( \rm x \) approaches 0 from either the positive or negative side, \( \rm x^2 \) approaches 0 through positive values, so \( \rm \frac{1}{x^2} \) approaches \( \rm +\infty \).

As the argument of the sine function approaches infinity, the value of \( \rm sin \left(\frac{1}{x^2}\right) \) oscillates between -1 and 1. It does not approach a single, specific value. For example, there are values of \( \rm x \) arbitrarily close to 0 for which \( \rm \frac{1}{x^2} \) is a multiple of \( \rm \pi \), making \( \rm sin \left(\frac{1}{x^2}\right) = 0 \). There are also values of \( \rm x \) arbitrarily close to 0 for which \( \rm \frac{1}{x^2} \) is of the form \( \rm 2n\pi + \frac{\pi}{2} \) or \( \rm 2n\pi + \frac{3\pi}{2} \), making \( \rm sin \left(\frac{1}{x^2}\right) = 1 \) or \( \rm -1 \) respectively.

Since the function values oscillate between -1 and 1 and do not settle on a single value as \( \rm x \to 0 \), the limit \( \rm \lim_{x \to 0} sin \left(\frac{1}{x^2}\right) \) does not exist.

For continuity at \( \rm x = 0 \), the limit must exist and be equal to \( \rm f(0) \). Since the limit does not exist, the function \( \rm f(x) \) cannot be continuous at \( \rm x = 0 \), regardless of the value assigned to \( \rm f(0) \).

Therefore, statement 1 is incorrect.

Statement 2: Continuity at \( \rm x = \frac{2}{\sqrt{\pi}} \)

The second statement claims that the function \( \rm f(x) \) is continuous at \( \rm x = \frac{2}{\sqrt{\pi}} \). For \( \rm x \ne 0 \), the function is defined as \( \rm f(x) = sin \left(\frac{1}{x^2}\right) \).

The point \( \rm x = \frac{2}{\sqrt{\pi}} \) is clearly not equal to 0. So, we can analyze the continuity of \( \rm f(x) \) at this point using the definition \( \rm f(x) = sin \left(\frac{1}{x^2}\right) \).

We can consider \( \rm f(x) \) as a composition of two functions:

• An inner function \( \rm g(x) = \frac{1}{x^2} \).
• An outer function \( \rm h(u) = sin(u) \).

So, \( \rm f(x) = h(g(x)) \).

Let's analyze the continuity of the inner function \( \rm g(x) = \frac{1}{x^2} \) at \( \rm x = \frac{2}{\sqrt{\pi}} \). This is a rational function. Rational functions are continuous at every point in their domain. The domain of \( \rm g(x) = \frac{1}{x^2} \) is all real numbers \( \rm x \) such that \( \rm x^2 \ne 0 \), which means \( \rm x \ne 0 \). Since \( \rm x = \frac{2}{\sqrt{\pi}} \ne 0 \), the function \( \rm g(x) \) is continuous at \( \rm x = \frac{2}{\sqrt{\pi}} \).

Now, let's consider the outer function \( \rm h(u) = sin(u) \). The sine function is continuous for all real numbers \( \rm u \).

A key property of continuous functions is that the composition of continuous functions is continuous. Since \( \rm g(x) \) is continuous at \( \rm x = \frac{2}{\sqrt{\pi}} \) and \( \rm h(u) \) is continuous at \( \rm u = g\left(\frac{2}{\sqrt{\pi}}\right) \), the composite function \( \rm f(x) = h(g(x)) \) is continuous at \( \rm x = \frac{2}{\sqrt{\pi}} \).

Alternatively, at any point \( \rm a \ne 0 \), we can evaluate the limit: \( \rm \lim_{x \to a} f(x) = \lim_{x \to a} sin \left(\frac{1}{x^2}\right) \) Since \( \rm \frac{1}{x^2} \) is continuous at \( \rm x = a \ne 0 \), \( \rm \lim_{x \to a} \frac{1}{x^2} = \frac{1}{a^2} \). Since the sine function is continuous, we can write: \( \rm \lim_{x \to a} sin \left(\frac{1}{x^2}\right) = sin \left(\lim_{x \to a} \frac{1}{x^2}\right) = sin \left(\frac{1}{a^2}\right) \) And \( \rm f(a) = sin \left(\frac{1}{a^2}\right) \) for \( \rm a \ne 0 \). So, \( \rm \lim_{x \to a} f(x) = f(a) \) for all \( \rm a \ne 0 \).

Specifically, at \( \rm x = \frac{2}{\sqrt{\pi}} \), the function is continuous.

Therefore, statement 2 is correct.

Conclusion

Statement 1 is incorrect because the limit of \( \rm f(x) \) as \( \rm x \to 0 \) does not exist.

Statement 2 is correct because \( \rm f(x) \) is a composition of functions that are continuous for all \( \rm x \ne 0 \), and \( \rm x = \frac{2}{\sqrt{\pi}} \) is a point where \( \rm x \ne 0 \).

Based on this analysis, only statement 2 is correct.