91Ó°ÊÓ

In calculus, understanding the concept of the "limit" of a function is essential. A limit portrays the value that a function "approaches" as the input (or variable) approaches a specific value. For the function given in the exercise, let's look at the behavior as \( x \rightarrow 0 \). By examining the expression \( x^n \sin\left(\frac{1}{x^2}\right) \), we see that although \( \sin\left(\frac{1}{x^2}\right) \) can oscillate between -1 and 1, the term \( x^n \) will approach 0 as \( x \) approaches 0, provided that \( n > 1 \).

Therefore, no matter what \( \sin\left(\frac{1}{x^2}\right) \) might be doing, the overall product will be getting closer and closer to 0. This allows the limit \( \lim _{x \rightarrow 0} f(x) = 0 \) to exist clearly for \( n > 1 \), confirming that the function stabilizes to this value at the point as \( x \) approaches 0. This behavior showcases how the limit examines the tendency of the function rather than focusing on the actual point values.

Continuity of a function at a particular point means there is no interruption, jump, or hole for the function at that point. Mathematically, a function \( f \) is continuous at a point \( x = a \) if \( \lim_{x \to a} f(x) = f(a) \).

For the exercise function, continuity at \( x = 0 \) implies checking this condition: does \( \lim_{x \to 0} f(x) \) equal \( f(0) \)? Given the prior steps, we know \( \lim_{x \to 0} f(x) = 0 \) and \( f(0) = 0 \). This equality confirms continuity since both the limit and the function value are the same at \( x = 0 \).

• Such assurance guarantees the absence of abrupt changes in the behavior of the function around this point.

For \( n > 1 \), this holds true, cementing the seamless transition of values and consistency required for continuity.

Differentiability involves understanding whether a function possesses a defined derivative at a particular point. A function is differentiable at a point if it is "smooth" around that point, essentially meaning it does not have sharp corners or cusps.

For our function, testing differentiability at \( x=0 \) requires computing the derivative \( f'(0) \). The derivative is obtained by the limit \( f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \). Substituting into this formula, we calculate \( \lim_{h \to 0} h^{n-1} \sin(\frac{1}{h^2}) \).

Since \( h^{n-1} \) approaches 0 when \( n > 1 \), the entire expression \( h^{n-1} \sin(\frac{1}{h^2}) \) also approaches 0. Thus, \( f'(0) \) exists, confirming differentiability for \( n > 1 \).

• Smoothness at \( x = 0 \) implies that not only does the function not break, but it also has a consistent slope represented by its derivative.

This highlights a step up from mere continuity, connoting a deeper layer of function behavior analysis.

A piecewise function is defined by different expressions or formulas over different interval segments of its domain. This format allows more flexibility, accommodating behaviors that could vary drastically over the domain.

In the given exercise, the function is defined piecewise as \( f(x) = x^n \sin\left(\frac{1}{x^2}\right) \) for \( x eq 0 \), and \( f(x) = 0 \) for \( x = 0 \).

• This ensures that the function adapts its form to smoothly transition or maintain a condition across points or intervals where the mathematical behavior might change.

Piecewise functions like this are invaluable in modeling real-world phenomena where simple algebraic functions are insufficient. They handily manage edge cases like those encountered in this exercise, where behavior at one point (\( x = 0 \)) differs radically from behavior just around it.