Evaluate $\lim_{k \to \infty} \int_0^1 \frac{(1-x)^k \cos(k/x)}{\sqrt{x}}dx$

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I want to evaluate $$\lim_{k \to \infty} \int_{(0,1)}\frac{(1-x)^k \cos(k/x)}{\sqrt{x}}dx$$

where the integral is Lebesgue-integral. Attempt:

Note first that for $x \in (0,1)$, we have $\lim_k \frac{(1-x)^k \cos(k/x)}{\sqrt{x}} = 0$. Also,

$$\left|\frac{(1-x)^k \cos(k/x)}{\sqrt{x}}\right| \leq 1/\sqrt{x}$$

and by monotone convergence theorem $$\int_{(0,1)}1/\sqrt{x}dx = \lim_n \int_{(1/n,1)} x^{-1/2}dx = \lim_n (2-2\sqrt{1/n})=2$$

so the dominated convergence theorem allows us to interchange integral and limit and we conclude that

$$\lim_{k \to \infty} \int_{(0,1)}\frac{(1-x)^k \cos(k/x)}{\sqrt{x}}dx = \int_{(0,1)}0 dx = 0$$

Is this correct?

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To simplify it a bit, note that the absolute value of the $k$th integral is bounded above by

$$\int_{(0,1)}\left |\frac{(1-x)^k \cos(k/x)}{\sqrt{x}}\right |\,dx \le \int_{(0,1)}\frac{(1-x)^k }{\sqrt{x}}\,dx$$

On the right the integrands are $\le 1/\sqrt x \in L^1.$ Furthermore these integrands $\to 0$ pointwise. The DCT then tells us the limit is $0.$