Question 1 Proof Let x 1 , x 2 be any two points of [ − 1 , 1 ] so that ∣ 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 ) ∣ = ∣ 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 choose δ = ϵ 3 , then ∣ f ( x 1 ) − f ( x 2 ) ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < δ ∴ f is uniformly continuous Question 2 Proof Let ϵ > 0 be given Let x 1 , x 2 be any two points of [ 1 , ∞ ) so that 1 ≤ x 1 and 1 ≤ x 2 ⇒ 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 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 . Now, for any δ > 0 ∃ m ∈ N such that 1 n < δ ∀ n > m Let x 1 = 1 2 m and x 2 = 1 m so that 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. \text{Question 1}
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\text{\textbf{Proof} } \text{Let } x_1, x_2 \text{ be any two points of } [-1,1] \text{ so that } \left| {{x_1}} \right| \le 1,\left| {{x_2}} \right| \le 1
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\text{Now. } \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|
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\hspace{95pt} = \left| {{x_1} - {x_2}} \right|\left| {x_1^2 + {x_1}{x_2} + x_2^2} \right|
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\hspace{95pt} \le \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)
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\hspace{95pt} \le 3\left| {{x_1} - {x_2}} \right| < \epsilon \text{ whenever } \left| {{x_1} - {x_2}} \right| < \frac{\epsilon }{3}
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\text{choose } \delta = \frac{\epsilon }{3}, \text{ then } \left| {f\left( {{x_1}} \right) - f\left( {{x_2}} \right)} \right| < \epsilon \text{ whenever } \left| {{x_1} - {x_2}} \right| < \delta
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\therefore f \text{ is uniformly continuous}
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\text{Question 2}
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\text{\textbf{Proof} } \text{Let } \epsilon > 0 \text{ be given}
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\text{Let } x_1, x_2 \text{ be any two points of } [1,\infty) \text{ so that } 1 \le {x_1} \text{ and } 1 \le {x_2}
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\hspace{30pt} \Rightarrow 2{x_1} - 1 \ge 1 \quad \text{and } 2{x_2} - 1
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\hspace{30pt} \Rightarrow \left| {2{x_1} - 1} \right| \ge 1 \quad \text{and } \left| {2{x_2} - 1} \right| \ge 1
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\hspace{30pt} \Rightarrow \frac{1}{{\left| {2{x_1} - 1} \right|}} \le 1 \quad \text{and } \Rightarrow \frac{1}{{\left| {2{x_2} - 1} \right|}} \le 1
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\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|
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\hspace{95pt} = \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|}}{{\left| {2{x_1} - 1} \right|\left| {2{x_2} - 1} \right|}}
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\hspace{95pt} \le 2\left| {{x_1} - {x_2}} \right| < \epsilon \text{ whenever } \left| {{x_1} - {x_2}} \right| < \frac{\epsilon }{2}
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\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
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\therefore f \text{ is uniformly continuous}
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\text{Question 3}
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\text{\textbf{Proof} } \text{Let } \epsilon > 0 \text{ be given}
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\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
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\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|
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\hspace{95pt} = \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
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\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
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\therefore f \text{ is uniformly continuous}
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\text{Question 4}
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\text{\textbf{Proof} } \text{Since } x \text{ is continuous on } I \text{and } x\ne0 \text{ in } I
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\therefore \frac{1}{x} \text{ is continuous}.
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\text{Now, for any } \delta>0 ~\exist m\in \mathbb{N} \text{ such that } \frac{1}{n}<\delta ~\forall n>m
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\hspace{30pt} \text{Let } {x_1} = \frac{1}{{2m}} \text{ and } {x_2} = \frac{1}{m} \text{ so that} x_1,x_2\in I
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\hspace{30pt} \text{and } \left| {{x_1} - {x_2}} \right| = \left| {\frac{1}{{2m}} - \frac{1}{m}} \right| = \frac{1}{{2m}} < \delta
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\text{ but } \left| {f\left( {{x_1}} \right) - f\left( {{x_2}} \right)} \right| = \left| {2m - m} \right| = m \text{ which cannot be less than } \epsilon > 0
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\therefore f \text{ is \textbf{not} uniformly continuous.} Question 1 Proof Let x 1 , x 2 be any two points of [ − 1 , 1 ] so that ∣ 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 ) ∣ ∣ = ∣ 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 = 3 ϵ , then ∣ f ( x 1 ) − f ( x 2 ) ∣ < ϵ whenever ∣ x 1 − x 2 ∣ < δ ∴ f is uniformly continuous Question 2 Proof Let ϵ > 0 be given Let x 1 , x 2 be any two points of [ 1 , ∞ ) so that 1 ≤ x 1 and 1 ≤ x 2 ⇒ 2 x 1 − 1 ≥ 1 and 2 x 2 − 1 ⇒ ∣ 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 = 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 s i n ( x 1 ) − x 2 s i n ( x 2 ) ∣ ∣ = ∣ ∣ x 1 x 2 x 2 s i n ( x 1 ) − x 1 s i n ( x 2 ) ∣ ∣ = ∣ x 1 ∣ ∣ x 2 ∣ ∣ x 2 s i n ( x 1 ) − x 1 s i n ( 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 such that n 1 < δ ∀ n > m Let x 1 = 2 m 1 and x 2 = m 1 so that 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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