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ta có $-\left| {x} \right|\leq xcos(\frac1{x^2})\leq \left| {x} \right|$
mà $\mathop {\lim }\limits_{x \to 0}-\left| {x} \right|=\mathop {\lim }\limits_{x \to 0}=\left| {x} \right|=0$
=>$\mathop {\lim }\limits_{x \to 0}xcos(\frac1{x^2})=0$(định lí kẹp)
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