Question 1
Proof Let x 1 , x 2 x_{1}, x_{2} x 1 , x 2 ∣ x 1 ∣ ≤ 1 , ∣ x 2 ∣ ≤ 1 \left|x_{1}\right| \leq 1,\left|x_{2}\right| \leq 1 ∣ x 1 ∣ ≤ 1 , ∣ x 2 ∣ ≤ 1
Now. ∣ f ( x 1 ) − f ( x 2 ) ∣ = ∣ x 1 3 − x 2 3 ∣ = ∣ ( x 1 − x 2 ) ( x 1 2 + x 1 x 2 + x 2 2 ) ∣ \left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|=\left|x_{1}^{3}-x_{2}^{3}\right|=\left|\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)\right| ∣ f ( x 1 ) − f ( x 2 ) ∣ = ∣ ∣ x 1 3 − x 2 3 ∣ ∣ = ∣ ∣ ( x 1 − x 2 ) ( x 1 2 + x 1 x 2 + x 2 2 ) ∣ ∣
= ∣ x 1 − x 2 ∣ ∣ x 1 2 + x 1 x 2 + x 2 2 ∣ ≤ ∣ x 1 − x 2 ∣ ( ∣ x 1 ∣ 2 + ∣ x 1 ∣ ∣ x 2 ∣ + ∣ x 2 ∣ 2 ) ≤ 3 ∣ x 1 − x 2 ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < ϵ 3 \begin{aligned}
&=\left|x_{1}-x_{2}\right|\left|x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right| \\
&\leq\left|x_{1}-x_{2}\right|\left(\left|x_{1}\right|^{2}+\left|x_{1}\right|\left|x_{2}\right|+\left|x_{2}\right|^{2}\right) \\
&\leq 3\left|x_{1}-x_{2}\right|<\epsilon \text { whenever }\left|x_{1}-x_{2}\right|<\frac{\epsilon}{3}
\end{aligned} = ∣ x 1 − x 2 ∣ ∣ ∣ x 1 2 + x 1 x 2 + x 2 2 ∣ ∣ ≤ ∣ x 1 − x 2 ∣ ( ∣ x 1 ∣ 2 + ∣ x 1 ∣ ∣ x 2 ∣ + ∣ x 2 ∣ 2 ) ≤ 3 ∣ x 1 − x 2 ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < 3 ϵ
choose δ = ϵ 3 \delta=\frac{\epsilon}{3} δ = 3 ϵ ∣ f ( x 1 ) − f ( x 2 ) ∣ < ϵ w h e n e v e r ∣ x 1 − x 2 ∣ < δ \left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\epsilon whenever \left|x_{1}-x_{2}\right|<\delta ∣ f ( x 1 ) − f ( x 2 ) ∣ < ϵ w h e n e v er ∣ x 1 − x 2 ∣ < δ
therefore f is uniformly continuous
Question 2
Proof Let ϵ > 0 \epsilon>0 ϵ > 0
Let x 1 , x 2 x_{1}, x_{2} x 1 , x 2 [ 1 , ∞ ) [1, \infty) [ 1 , ∞ ) 1 ≤ x 1 1 \leq x_{1} 1 ≤ x 1 1 ≤ x 2 1 \leq x_{2} 1 ≤ x 2
⇒ 2 x 1 − 1 ≥ 1 and 2 x 2 − 1 \Rightarrow 2 x_{1}-1 \geq 1 \text { and } 2 x_{2}-1 ⇒ 2 x 1 − 1 ≥ 1 and 2 x 2 − 1
⇒ ∣ 2 x 1 − 1 ∣ ≥ 1 and ∣ 2 x 2 − 1 ∣ ≥ 1 ⇒ 1 ∣ 2 x 1 − 1 ∣ ≤ 1 and ⇒ 1 2 x 2 − 1 ≤ 1 Now ∣ f ( x 1 ) − f ( x 2 ) ∣ = ∣ 2 x 1 2 x 1 − 1 − 2 x 2 2 x 2 − 1 ∣ = ∣ 2 x 1 ( 2 x 2 − 1 ) − 2 x 2 ( 2 x 1 − 1 ) ( 2 x 1 − 1 ) ( 2 x 2 − 1 ) ∣ = 2 ∣ x 1 − x 2 ∣ 2 x 1 − 1 ∣ ∣ 2 x 2 − 1 ∣ ≤ 2 ∣ x 1 − x 2 ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < ϵ 2 \begin{aligned}
&\quad \Rightarrow\left|2 x_{1}-1\right| \geq 1 \quad \text { and }\left|2 x_{2}-1\right| \geq 1 \\
&\quad \Rightarrow \frac{1}{\left|2 x_{1}-1\right|} \leq 1 \quad \text { and } \quad \Rightarrow \frac{1}{2 x_{2}-1} \leq 1 \\
&\text { Now }\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|=\left|\frac{2 x_{1}}{2 x_{1}-1}-\frac{2 x_{2}}{2 x_{2}-1}\right| \\
&\quad=\left|\frac{2 x_{1}\left(2 x_{2}-1\right)-2 x_{2}\left(2 x_{1}-1\right)}{\left(2 x_{1}-1\right)\left(2 x_{2}-1\right)}\right|=\frac{2\left|x_{1}-x_{2}\right|}{2 x_{1}-1|| 2 x_{2}-1 \mid} \\
&\quad \leq 2\left|x_{1}-x_{2}\right|<\epsilon \text { whenever }\left|x_{1}-x_{2}\right|<\frac{\epsilon}{2}
\end{aligned} ⇒ ∣ 2 x 1 − 1 ∣ ≥ 1 and ∣ 2 x 2 − 1 ∣ ≥ 1 ⇒ ∣ 2 x 1 − 1 ∣ 1 ≤ 1 and ⇒ 2 x 2 − 1 1 ≤ 1 Now ∣ f ( x 1 ) − f ( x 2 ) ∣ = ∣ ∣ 2 x 1 − 1 2 x 1 − 2 x 2 − 1 2 x 2 ∣ ∣ = ∣ ∣ ( 2 x 1 − 1 ) ( 2 x 2 − 1 ) 2 x 1 ( 2 x 2 − 1 ) − 2 x 2 ( 2 x 1 − 1 ) ∣ ∣ = 2 x 1 − 1∣∣2 x 2 − 1 ∣ 2 ∣ x 1 − x 2 ∣ ≤ 2 ∣ x 1 − x 2 ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < 2 ϵ
choose δ = ϵ 2 , then ∣ f ( x 1 ) − f ( x 2 ) ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < δ ∴ f is uniformly continuous Question 3 Proof Let ϵ > 0 be given Let x 1 , x 2 be any two points of ( 0 , 1 ) so that 0 < x 1 < 1 and 0 < x 2 < 1 Now ∣ f ( x 1 ) − f ( x 2 ) ∣ = ∣ sin ( x 1 ) x 1 − sin ( x 2 ) x 2 ∣ = ∣ x 2 sin ( x 1 ) − x 1 sin ( x 2 ) x 1 x 2 ∣ = ∣ x 2 sin ( x 1 ) − x 1 sin ( x 2 ) ∣ ∣ x 1 ∣ ∣ x 2 ∣ < ∣ x 2 x 1 − x 1 x 2 ∣ < ∣ x 1 − x 2 ∣ < ϵ take δ = ϵ , then ∣ f ( x 1 ) − f ( x 2 ) ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < δ ∴ f is uniformly continuous Question 4 Proof Since x is continuous on I and x ≠ 0 in I ∴ 1 x is continuous. \begin{aligned}
&\text { choose } \delta=\frac{\epsilon}{2}, \text { then }\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\epsilon \text { whenever }\left|x_{1}-x_{2}\right|<\delta \\
&\therefore f \text { is uniformly continuous } \\
&\text { Question } 3 \\
&\text { Proof Let } \epsilon>0 \text { be given } \\
&\text { Let } x_{1}, x_{2} \text { be any two points of }(0,1) \text { so that } 0<x_{1}<1 \text { and } 0<x_{2}<1 \\
&\text { Now }\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|=\left|\frac{\sin \left(x_{1}\right)}{x_{1}}-\frac{\sin \left(x_{2}\right)}{x_{2}}\right|=\left|\frac{x_{2} \sin \left(x_{1}\right)-x_{1} \sin \left(x_{2}\right)}{x_{1} x_{2}}\right| \\
&\qquad=\frac{\left|x_{2} \sin \left(x_{1}\right)-x_{1} \sin \left(x_{2}\right)\right|}{\left|x_{1}\right|\left|x_{2}\right|}<\left|x_{2} x_{1}-x_{1} x_{2}\right|<\left|x_{1}-x_{2}\right|<\epsilon \\
&\text { take } \delta=\epsilon, \text { then }\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\epsilon \text { whenever }\left|x_{1}-x_{2}\right|<\delta \\
&\therefore f \text { is uniformly continuous } \\
&\text { Question } 4 \\
&\text { Proof Since } x \text { is continuous on } I \text { and } x \neq 0 \text { in } I \\
&\therefore \frac{1}{x} \text { is continuous. }
\end{aligned} choose δ = 2 ϵ , then ∣ f ( x 1 ) − f ( x 2 ) ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < δ ∴ f is uniformly continuous Question 3 Proof Let ϵ > 0 be given Let x 1 , x 2 be any two points of ( 0 , 1 ) so that 0 < x 1 < 1 and 0 < x 2 < 1 Now ∣ f ( x 1 ) − f ( x 2 ) ∣ = ∣ ∣ x 1 sin ( x 1 ) − x 2 sin ( x 2 ) ∣ ∣ = ∣ ∣ x 1 x 2 x 2 sin ( x 1 ) − x 1 sin ( x 2 ) ∣ ∣ = ∣ x 1 ∣ ∣ x 2 ∣ ∣ x 2 sin ( x 1 ) − x 1 sin ( x 2 ) ∣ < ∣ x 2 x 1 − x 1 x 2 ∣ < ∣ x 1 − x 2 ∣ < ϵ take δ = ϵ , then ∣ f ( x 1 ) − f ( x 2 ) ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < δ ∴ f is uniformly continuous Question 4 Proof Since x is continuous on I and x = 0 in I ∴ x 1 is continuous.
Now, for any δ > 0 ∃ m ∈ N s u c h t h a t 1 n < δ ∀ n > m Let x 1 = 1 2 m a n d x 2 = 1 m s o t h a t x 1 , x 2 ∈ I and ∣ x 1 − x 2 ∣ = ∣ 1 2 m − 1 m ∣ = 1 2 m < δ but ∣ f ( x 1 ) − f ( x 2 ) ∣ = ∣ 2 m − m ∣ = m which cannot be less than ϵ > 0 ∴ f is not uniformly continuous. \begin{aligned}
& \text{Now, for any} \delta>0 \exists m \in \mathbb{N} such that \frac{1}{n}<\delta \forall n>m\\
& \text{Let} x_{1}=\frac{1}{2 m} and x_{2}=\frac{1}{m} so that x_{1}, x_{2} \in I\\
& \text{and} \left|x_{1}-x_{2}\right|=\left|\frac{1}{2 m}-\frac{1}{m}\right|=\frac{1}{2 m}<\delta\\
& \text{but} \left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|=|2 m-m|=m \text{which cannot be less than} \epsilon>0\\
&\therefore \text{f is not uniformly continuous.}\\
\end{aligned} Now, for any δ > 0∃ m ∈ N s u c h t ha t n 1 < δ ∀ n > m Let x 1 = 2 m 1 an d x 2 = m 1 so t ha t x 1 , x 2 ∈ I and ∣ x 1 − x 2 ∣ = ∣ ∣ 2 m 1 − m 1 ∣ ∣ = 2 m 1 < δ but ∣ f ( x 1 ) − f ( x 2 ) ∣ = ∣2 m − m ∣ = m which cannot be less than ϵ > 0 ∴ f is not uniformly continuous.
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