Given ε > 0 \varepsilon>0 ε > 0 , choose δ = ε \delta=\sqrt{\varepsilon} δ = ε
Now, if x x x is chosen so that 0 < ∣ x − 0 ∣ < δ 0<|x-0|<\delta 0 < ∣ x − 0∣ < δ ,
then
∣ x 2 cos ( 1 x ) − 0 ∣ = ∣ x 2 ∣ ∣ cos ( 1 x ) ∣ = x 2 ∣ cos ( 1 x ) ∣ ≤ x 2 ( 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} ∣ ∣ x 2 cos ( x 1 ) − 0 ∣ ∣ = ∣ ∣ x 2 ∣ ∣ ∣ ∣ cos ( x 1 ) ∣ ∣ = x 2 ∣ ∣ cos ( x 1 ) ∣ ∣ ≤ x 2 ( 1 ) < ( ε ) 2 = ε
That is: if 0 < ∣ x − 0 ∣ < δ 0<|x-0|<\delta 0 < ∣ x − 0∣ < δ , then ∣ x 2 cos ( 1 x ) − 0 ∣ < ε \left|x^{2} \cos \left(\frac{1}{x}\right)-0\right|<\varepsilon ∣ ∣ x 2 cos ( x 1 ) − 0 ∣ ∣ < ε
So, by definition of limit, lim x → 0 ( x 2 cos ( 1 x ) ) = 0. \lim _{x \rightarrow 0}\left(x^{2} \cos \left(\frac{1}{x}\right)\right)=0. lim x → 0 ( x 2 cos ( x 1 ) ) = 0.
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