Đáp án:
$ 9)A$
$17)B$
Giải thích các bước giải:
$9)A.\displaystyle\lim_{x \to +\infty} x \sin \dfrac{1}{x}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{\sin \dfrac{1}{x}}{\dfrac{1}{x}}\\ =1(x \to +\infty \Rightarrow \dfrac{1}{x} \to 0) \ne +\infty\\ \Rightarrow S\\ B. \displaystyle\lim_{x \to 0} \dfrac{\cos x}{x}\\ =\dfrac{1}{0}\\ =+\infty\\ \Rightarrow Đ\\ C. \displaystyle\lim_{x \to +\infty} x \cos \dfrac{1}{x}\\ =\infty \cos 0\\ =\infty\\ \Rightarrow Đ\\ D.\displaystyle\lim_{x \to +\infty} x \sin \dfrac{1}{x}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{\sin \dfrac{1}{x}}{\dfrac{1}{x}}\\ =1(x \to +\infty \Rightarrow \dfrac{1}{x} \to 0) \\ \Rightarrow Đ\\ 17)f(x)=(x\sin \alpha +\cos \alpha)(x\cos \alpha -\sin \alpha)\\ f'(x)=(x\sin \alpha +\cos \alpha)'(x\cos \alpha -\sin \alpha)+(x\sin \alpha +\cos \alpha)(x\cos \alpha -\sin \alpha)'\\ =(x'\sin \alpha +x\sin \alpha' )(x\cos \alpha -\sin \alpha)+(x\sin \alpha +\cos \alpha)(x'\cos \alpha+x\cos \alpha' )\\ =\sin \alpha(x\cos \alpha -\sin \alpha)+\cos \alpha(x\sin \alpha +\cos \alpha)\\ = x\sin \alpha \cos \alpha -\sin^2 \alpha+x\cos \alpha\sin \alpha +\cos^2 \alpha\\ =2x\sin \alpha \cos \alpha+\cos^2 \alpha -\sin^2\alpha\\ =x\sin 2\alpha+\cos 2\alpha\\ \Rightarrow B$
