Define f ( x ) = 3 x ∀ x ∈ [ 2 , 4 ] . f(x)=\frac{3}{x} \forall x \in[2,4]. f ( x ) = x 3 ∀ x ∈ [ 2 , 4 ] .
Then f is continuous and bounded function on [2,4].
For any δ > 0 \delta>0 δ > 0 , we can find m ∈ N m \in \mathbb{N} m ∈ N such that 1 n < δ \frac{1}{n}<\delta n 1 < δ for every n ≥ m n \geq m n ≥ m
Let x 1 = 3 m , x 2 = 3 2 m x_{1}=\frac{3}{m}, x_{2}=\frac{3}{2 m} x 1 = m 3 , x 2 = 2 m 3 , so that x 1 , x 2 ∈ [ 2 , 4 ] x_{1}, x_{2} \in[2,4] x 1 , x 2 ∈ [ 2 , 4 ]
Consider ∣ x 2 − x 1 ∣ = ∣ 3 2 m − 3 m ∣ = 3 2 m > δ 2 > δ \left|x_{2}-x_{1}\right|=\left|\frac{3}{2 m}-\frac{3}{m}\right|=\frac{3}{2 m}>\frac{\delta}{2}>\delta ∣ x 2 − x 1 ∣ = ∣ ∣ 2 m 3 − m 3 ∣ ∣ = 2 m 3 > 2 δ > δ
and ∣ f ( x 2 ) − f ( x 1 ) ∣ = ∣ 2 m 3 − m 3 ∣ = m 3 \left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|=|\frac{2 m}{3}-\frac{m}{3}|=\frac{m}{3} ∣ f ( x 2 ) − f ( x 1 ) ∣ = ∣ 3 2 m − 3 m ∣ = 3 m
Therefore if we choose ε > 0 \varepsilon>0 ε > 0 such that ε > m \varepsilon>m ε > m
then we get ∣ x 2 − x 1 ∣ > δ \left|x_{2}-x_{1}\right|>\delta ∣ x 2 − x 1 ∣ > δ and ∣ f ( x 2 ) − f ( x 1 ) ∣ < ε \left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|<\varepsilon ∣ f ( x 2 ) − f ( x 1 ) ∣ < ε
Hence f is uniformly continuous on [2,4].
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