[x^2*sin(1/x^2)]/x的X趋于0的极限,为什么不能用sin(1/x^2)~1/x^2带入.
举一反三
- 17e0b849d3a4a3b.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]的实验命令为（ ）. A: syms x; f=diff((1+sin(x)^2)/cos(x),1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2 B: f=diff((1+sinx^2)/cosx,1)f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;f=diff((1+sinx^2)/cosx,1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2
- 求极限\( \lim \limits_{x \to 0} { { {x^2}\sin {1 \over x}} \over {\sin x}}{\rm{ = }}\)______
- \( \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = \)______ 。
- 求函数[img=107x38]17da6537b12a2e0.png[/img]的导数； ( ) A: 2*x*sin(1/x) - sin(1/x) B: 2xsin(1/x) - cos(1/x) C: 2*x*sin(1/x) - cos(1/x) D: 2*x*cos(1/x) - cos(1/x)
- 函数\(y = \sin {1 \over x}\)的导数为（ ）. A: \({1 \over { { x^2}}}\sin {1 \over x}\) B: \( - {1 \over { { x^2}}}\sin {1 \over x}\) C: \( - {1 \over { { x^2}}}\cos {1 \over x}\) D: \({1 \over { { x^2}}}\cos {1 \over x}\)