91Ó°ÊÓ

To determine if a function is differentiable at a given point, it must first be continuous at that point. Let's look at the piecewise function given here:

• For \( x eq 0 \), the function is \( f(x) = e^{-1/x^2} \).
• At \( x = 0 \), the function is \( f(x) = 0 \).

Continuity at a point means that the left-hand limit and right-hand limit are equal to the function's value at that point.Let's check the continuity of \( f(x) \) at \( x = 0 \).1. **Limit evaluation of \( f(x) \) as \( x \to 0 \):** \[ \lim_{x \to 0} e^{-1/x^2} = 0 \]2. **Function value at 0:** \[ f(0) = 0 \]Both the limit and the function value at \( x = 0 \) are 0. Hence, \( f(x) \) is continuous at \( x = 0 \). This is a must-have requirement for differentiability at that point.

Evaluating limits is essential in understanding the behavior of functions as they approach a particular point. For the given function, we need to evaluate the limit to check continuity and to find the derivative.The magical aspect of functions like \( f(x) = e^{-1/x^2} \) is that they converge rapidly to 0 as \( x \to 0 \). In mathematical terms:

• To find the limit: \[ \lim_{x \to 0} e^{-1/x^2} = 0 \]
• This tells us that regardless of the path \( x \) takes to approach 0, \( e^{-1/x^2} \) stays glued to 0.

This amazing characteristic assists not only in establishing continuity but also plays a pivotal role in determining the derivative's behavior at that exceptional point.

The derivative of a function at a point provides the slope of the tangent to the function's graph at that point. The definition of the derivative at any point \( a \) is the limit of the difference quotient:\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]Specifically, we want to find \( f'(0) \) for our function, which requires us to compute this limit:\[ f'(0) = \lim_{h \to 0} \frac{e^{-1/h^2} - 0}{h} = \lim_{h \to 0} \frac{e^{-1/h^2}}{h} \]Here, as the very small \( h \) approaches zero, the value \( e^{-1/h^2} \) trends swiftly towards zero *much* faster than the denominator \( h \) becomes insignificant. This implies:

• In lopsided terms of dominance, \( e^{-1/h^2} \) becomes the underdog, barely contributing any heft.
• Thus, \( \lim_{h \to 0} \frac{e^{-1/h^2}}{h} = 0 \).

So, \( f'(0) \) emerges as 0, reflecting the smooth flat tangent mirroring through that narrow bridge of 0.