91Ó°ÊÓ

To check the continuity of the function \(f_n(x)\), we need to show that the function is continuous at \(x = 0\) as well as for \(x \neq 0\). We can see that \(f_n(x) = x^n \cos\frac{1}{x}\) is continuous for \(x \neq 0\) for all n since it is a product of continuous functions. Now let's check the continuity at \(x = 0\): For \(x = 0\), we have \(f_n(0) = 0\). So, we need to find the limit of the function as \(x\) approaches \(0\) for each \(n\). If the limit exists and is equal to the function value at \(x=0\), then the function is continuous at \(x=0\). $$ \lim_{x \to 0} x^n \cos\frac{1}{x} $$ For any \(n \geq 1\), we know that \(\lim_{x \to 0} x^n = 0\) and also \(-1 \leq \cos\frac{1}{x} \leq 1\). By applying the squeeze theorem, we can see that: $$ \lim_{x \to 0} x^n \cos\frac{1}{x} = 0 $$ Since the limit exists and is equal to the function value at \(x=0\), the function \(f_n(x)\) is continuous for \(n=1,2,3\).

Now, we will differentiate the function \(f_n(x)\) for \(n=1, 2, 3\). We will use the product rule for differentiation: 1. For \(n = 1\): $$ f_1'(x) = (x\cos\frac{1}{x})' = \cos\frac{1}{x} - \frac{1}{x^2}\sin\frac{1}{x} $$ 2. For \(n = 2\): $$ f_2'(x) = (x^2\cos\frac{1}{x})' = 2x\cos\frac{1}{x} - \frac{x^2}{x^3}\sin\frac{1}{x} = 2x\cos\frac{1}{x} - \frac{1}{x}\sin\frac{1}{x} $$ 3. For \(n = 3\): $$ f_3'(x) = (x^3\cos\frac{1}{x})' = 3x^2\cos\frac{1}{x} - \frac{x^3}{x^4}\sin\frac{1}{x} = 3x^2\cos\frac{1}{x} - \frac{1}{x^2}\sin\frac{1}{x} $$ Now we will check the differentiability of the function.

A function is differentiable at a point if its derivative exists at that point. We have already found the derivatives for \(f_1'(x)\), \(f_2'(x)\), and \(f_3'(x)\) for \(x \neq 0\). Now we need to check the differentiability at \(x=0\) for each function: 1. For \(n = 1\): $$ \lim_{x \to 0} f_1'(x) = \lim_{x \to 0} (\cos\frac{1}{x} - \frac{1}{x^2}\sin\frac{1}{x}) $$ Since the second term of the expression does not have a limit as x approaches 0, so the limit does not exist. Therefore, \(f_1(x)\) is not differentiable. 2. For \(n = 2\): $$ \lim_{x \to 0} f_2'(x) = \lim_{x \to 0} (2x\cos\frac{1}{x} - \frac{1}{x}\sin\frac{1}{x}) $$ This limit does not exist as well. Therefore, \(f_2(x)\) is not differentiable. 3. For \(n = 3\): $$ \lim_{x \to 0} f_3'(x) = \lim_{x \to 0} (3x^2\cos\frac{1}{x} - \frac{1}{x^2}\sin\frac{1}{x}) $$ Using the squeeze theorem, we find that this limit exists and equals 0. Thus, \(f_3(x)\) is differentiable at \(x=0\), and it is differentiable for all x in the domain of the function.

Since \(f_3(x)\) is differentiable for all x in the domain of the function, it has continuous derivatives as well. Thus \(f_3(x)\) is continuously differentiable. To summarize, for the given function \(f_n(x)\): 1. Continuous for n = 1, 2, 3 2. Differentiable only for n = 3 3. Continuously differentiable only for n = 3