find the limit lim x→0 x^2 cos 1/x.
Given ε>0\varepsilon>0ε>0 , choose δ=ε\delta=\sqrt{\varepsilon}δ=ε
Now, if xxx is chosen so that 0<∣x−0∣<δ0<|x-0|<\delta0<∣x−0∣<δ ,
then
∣x2cos(1x)−0∣=∣x2∣∣cos(1x)∣=x2∣cos(1x)∣≤x2(1)<(ε)2=ε\begin{aligned} \left|x^{2} \cos \left(\frac{1}{x}\right)-0\right| &=\left|x^{2}\right|\left|\cos \left(\frac{1}{x}\right)\right| \\ &=x^{2}\left|\cos \left(\frac{1}{x}\right)\right| \\ & \leq x^{2}(1) \\ &<(\sqrt{\varepsilon})^{2}=\varepsilon \end{aligned}∣∣x2cos(x1)−0∣∣=∣∣x2∣∣∣∣cos(x1)∣∣=x2∣∣cos(x1)∣∣≤x2(1)<(ε)2=ε
That is: if 0<∣x−0∣<δ0<|x-0|<\delta0<∣x−0∣<δ , then ∣x2cos(1x)−0∣<ε\left|x^{2} \cos \left(\frac{1}{x}\right)-0\right|<\varepsilon∣∣x2cos(x1)−0∣∣<ε
So, by definition of limit, limx→0(x2cos(1x))=0.\lim _{x \rightarrow 0}\left(x^{2} \cos \left(\frac{1}{x}\right)\right)=0.limx→0(x2cos(x1))=0.
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