91Ó°ÊÓ

Firstly, let's focus on the pointwise convergence of the sequence of functions \(f_k(x)\). By analyzing the function, we can see that as \(k \rightarrow \infty\): \[ f_k(x) = \frac{(1-x)^k\cos\left(\frac{k}{x}\right)}{\sqrt{x}} \] As \(k \rightarrow \infty\), the term \((1-x)^k\) approaches \(0\) for \(x \in (0, 1]\) due to the exponential decay. Since \(\cos\left(\frac{k}{x}\right)\) is always bounded between -1 and 1 and \(\frac{1}{\sqrt{x}}\) is always positive for \(x \in (0, 1]\), the whole expression \(f_k(x)\) converges pointwise to a function \(f(x)\) such that \(f(x) = 0\) for \(x \in (0, 1]\).

Next, we need to find an integrable function \(g(x)\) that dominates all functions in the sequence of functions \(f_k(x)\) for \(x \in (0, 1]\). Since \(|\cos(z)| \leq 1\), we can bound \(f_k(x)\) as follows: \[ |f_k(x)| = \left|\frac{(1-x)^k\cos\left(\frac{k}{x}\right)}{\sqrt{x}}\right| \leq \frac{(1-x)^k}{\sqrt{x}} \] Now, let's consider the function \(g(x) = \frac{1}{\sqrt{x}}\). Notice that for any \(k\) and any \(x \in (0, 1]\), we have: \[ |f_k(x)| \leq \frac{(1-x)^k}{\sqrt{x}} \leq \frac{1}{\sqrt{x}} = g(x) \] The function \(g(x)\) is integrable on the interval \((0, 1]\) since its integral converges: \[ \int_0^1 \frac{1}{\sqrt{x}}\, dx = \left[2\sqrt{x}\right]_0^1 = 2 \] Since \(g(x)\) is integrable and it dominates the sequence of functions \(|f_k(x)|\), we can use the Dominated Convergence Theorem.

By the Dominated Convergence Theorem, we have: \[ \lim_{k \rightarrow \infty} \int_0^1 f_k(x)\, dx = \int_0^1 \lim_{k \rightarrow \infty} f_k(x)\, dx = \int_0^1 f(x)\, dx \] Since we determined in Step 1 that the pointwise limit function \(f(x) = 0\), we can simplify the integral: \[ \int_0^1 f(x)\, dx = \int_0^1 0\, dx = 0 \] Therefore, the limit of the integral of the sequence of functions \(f_k(x)\) as \(k \rightarrow \infty\) is equal to 0: \[ \lim_{k \rightarrow \infty} \int_{0}^{1} f_k(x)\, dx = 0 \]