Answer to Question #264657 in Real Analysis for Dhruv bartwal

Question

The function f: [ 2,4] to R , defined by f(x)= 3/x is uniformly continuous on its domain.

True or false with full explanation

Answer 1

Define $f(x)=\frac{3}{x} \forall x \in[2,4].$

Then f is continuous and bounded function on [2,4].

For any $\delta>0$ , we can find $m \in \mathbb{N}$ such that $\frac{1}{n}<\delta$ for every $n \geq m$

Let $x_{1}=\frac{3}{m}, x_{2}=\frac{3}{2 m}$ , so that $x_{1}, x_{2} \in[2,4]$

Consider $\left|x_{2}-x_{1}\right|=\left|\frac{3}{2 m}-\frac{3}{m}\right|=\frac{3}{2 m}>\frac{\delta}{2}>\delta$

and $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|=|\frac{2 m}{3}-\frac{m}{3}|=\frac{m}{3}$

Therefore if we choose $\varepsilon>0$ such that $\varepsilon>m$

then we get $\left|x_{2}-x_{1}\right|>\delta$ and $\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|<\varepsilon$

Hence f is uniformly continuous on [2,4].