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Differentiability is a property that describes whether a function has a derivative at every point in its domain. A function is considered differentiable at a specific point if the derivative, which is the slope of the tangent line at that point, exists. This essentially means that the function must be smooth, or without any sharp changes, at that point.

To check if a function like \( f(x) \) is differentiable at a point (for example, \( x=0 \)), we need to examine the limit:

• \( \lim_{{h \to 0}} \frac{f(h) - f(0)}{h} \)

If this limit exists and is finite, then the function is differentiable at that point. In our case, the continuity check also plays a role because for differentiability, the function must first be continuous.

In Step 3 of the solution, we used the squeeze theorem to show that the limit exists, proving differentiability at \( x=0 \). Differentiability may not imply continuity of the derivative function, as observed in Step 4, where \( f'(x) \) is not continuous at \( x=0 \).

The squeeze theorem is a powerful mathematical tool used to find the limit of a function that is "squeezed" between two other functions. This theorem is particularly useful when a function is caught between two functions whose limits at a certain point are known.

For some functions where direct evaluation of limits is difficult or impossible, the squeeze theorem helps provide clarity. If each bounding function approaches the same limit as \( x \) approaches a specific value, the squeezed function will also approach this limit. The following inequality is used to apply the squeeze theorem:

• "If \( g(x) \leq f(x) \leq h(x) \) for all \( x \) near \( a \), and \( \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \), then \( \lim_{x \to a} f(x) = L \)."

In the exercise, we squeezed \( x^2 \cos \left(\frac{2}{x}\right) \) between \( -x^2 \) and \( x^2 \), both of which approach 0 as \( x \to 0 \). This results in \( f(x) \) also approaching 0, demonstrating the continuity of the function at this point.

The product rule is essential when differentiating functions that are products of two or more functions. This rule states that the derivative of a product of two functions \( u(x) \) and \( v(x) \) is a combination of their derivatives. Mathematically, it is represented as:

• \( (uv)' = u'v + uv' \)

This rule is crucial in calculating derivatives in complex calculus problems, like when dealing with compositions of multiple functions.

In Step 2 of the solution, the product rule was used to find the derivative \( f'(x) \) for \( x eq 0 \). By defining \( u = x^2 \) and \( v = \cos \left(\frac{2}{x}\right) \), their respective derivatives \( u' = 2x \) and \( v' = \sin \left(\frac{2}{x}\right) \left(-\frac{2}{x^2}\right) \) are plugged into the product rule formula, resulting in the derivative expression for the given function.

Trigonometric limits are another crucial aspect when dealing with functions involving trigonometric expressions. These limits deal with the behavior of trigonometric functions as they approach a particular point, often 0.

Some classic limits to remember include:

• \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
• \( \lim_{x \to 0} \cos x = 1 \)

These fundamental limits help evaluate the behavior of more complex trigonometric expressions within functions, especially where direct substitution might lead to indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

In the problem, recognizing how the trigonometric function \( \cos \left(\frac{2}{x}\right) \) behaves was central to determining the limits using the squeeze theorem, demonstrating both the role of basic trigonometric limits and more complex ones in the context of continuity and differentiability.