Question (Uniform Continuity): Show that 1/x is continuous on (0, ∞) but not uniformly continuous.
Answer :
Continuous function : A function f(x) is called continuous at a point p if for any given ε > 0 there exists a δ > 0 such that |x − p| < δ implies |f(x) − f(p)| <ε
To show that 1/x is continuous on (0, ∞) :
Let p be a point in (0, ∞) clearly p>0
Calculating f(p) =1/p
To prove \[ limit_{x->p}f(x)=f(p)\]
Taking LHS
\[ limit_{x->p} \frac{1}{x} \]
Left hand limit
\[ limit_{h->0} \frac{1}{p-h} = \frac{1}{p} \]
Right hand limit
\[ limit_{h->0} \frac{1}{p+h} = \frac{1}{p} \]
Since Left hand limit =Right hand Limit hence
\[ => limit_{x->p} \frac{1}{x} \]
and
\[ limit_{x->p} \frac{1}{x}= f(p) =\frac{1}{p}\]
Therefore 1/x is continuous on (0, ∞) at x = p.
Uniformly Continuous function: A function f(x) is called uniformly continuous on (a, b) if for a given ε > 0 then there exists a δ > 0 such that |x − y| < δ implies |f(x) − f(y)| <ε .
To show that 1/x is not uniformly continuous on (0, ∞):
Proof by contradiction, Let’ s suppose that 1/x is not uniformly continuous on (0, ∞)
Let ε=1/3 is given then there exists a δ such that
|x − y| < δ => |f(x) − f(y)| <ε
Let K>1 be a large positive number such that 1/K < δ.
Let choosing \[ x = \frac{1}{K}, y =\frac{1}{K+1}\]
Then it clear that
\[ |x − y| =|\frac{1}{K} − \frac{1}{K+1}|=|\frac{K+1-K}{K(K+1)} | =|\frac{1}{K(K+1)} |< δ \]
But if we calculate \[ |f(x) − f(y)| =|\frac{1}{\frac{1}{K}}-\frac{1}{\frac{1}{K+1}}|=|M-(M+1)|=1>\frac{1}{3}=ε \]
Which is false
Therefore our assumption is wrong and hence 1/x is not uniformly continuous on (0, ∞)