请帮忙指出下列用等价无穷小替换求极限错在哪儿

$\lim\limits_{x \to 0}$$\cfrac{sin({x}^{2}sin\cfrac{1}{x})}{x}$=$\lim\limits_{x \to 0}$$\cfrac{{x}^{2}sin\cfrac{1}{x}}{x}$=$\lim\limits_{x \to 0}$$xsin\cfrac{1}{x}$=0
这样子错了吗?
已邀请:

poorich

赞同来自:

我就是来试试公式

$\lim\limits_{x \to 0} \cfrac{{\sin ({x^2}\sin \cfrac{1}{x})}}{x}
=\lim\limits_{x \to 0}
\cfrac{{\sin ({x^2}\sin \cfrac{1}{x})}}
{{{x^2}\sin \frac{1}{x}}}
\cfrac{{{x^2}\sin \cfrac{1}{x}}}{x}
\xlongequal[]{\lim\limits_{x \to 0} {x^2}\sin \cfrac{1}{x} = 0}
\lim\limits_{x \to 0} \cfrac{{{x^2}\sin \frac{1}{x}}}{x} =
\lim\limits_{x \to 0} x\sin \cfrac{1}{x}=0$

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