Math 131AReal AnalysisQuiz 3Instructions: You have 30 minutes to complete the quiz. There are two problems, worth atotal of 20 points. You may not use any books or notes. Partial credit will be given for progresstoward correct proofs. Write your solutions in the space below the questions. If you need morespace use the back of the page. Do not forget to write your name and UID in the space below.Name:Student ID number:Question Points Score1 102 10Total: 20Problem 1.Suppose s : [0, 1] → R, t : [1, 2] → R, and u : [2, 4] → R are continuous functions.Suppose that s(1) = t(1) and t(2) 6= u(2).Let f : [0, 2] → R be defined byf(x) =(s(x) if x ∈ [0, 1]t(x) if x ∈ [1, 2].Let g : [1, 4) → R be defined byg(x) =(t(x) if x ∈ [1, 2]u(x) if x ∈ (2, 4).(a) [6pts.] Prove that f is continuous at 1 using the -δ definition.Solution: Let  > 0.Since s is continuous at 1, there is a δs∈ (0, 1) such that1 − δs< x ≤ 1 =⇒ |s(x) − s(1)| < .Since t is continuous at 1, there is a δt∈ (0, 1) such that1 ≤ x < 1 + δt=⇒ |t(x) − t(1)| < .Let δ = min{δs, δt}. Then δ > 0.Suppose x ∈ [0, 2] and |x − 1| < δ. Then either• 1 − δ < x ≤ 1, in which case 1 − δs< x ≤ 1, so that|f(x) − f(1)| = |s(x) − s(1)| < ;or• 1 ≤ x < 1 + δ, in which case 1 ≤ x < 1 + δt, so that|f(x) − f(1)| = |t(x) − t(1)| < .(b) [4pts.] Prove that g is NOT continuous at 2 using the sequence definition.Solution: Consider the sequence (xn)∞n=1defined by xn= 2 +1n.We have limn→∞xn= 2.However, using that xn∈ (2, 4) for each n ∈ N and the definition of g, the factthat u is continuous at 2, and the definition of g again, we see thatlimn→∞g(xn) = limn→∞u(xn) = u(2) 6= t(2) = g(2).Problem 2.(a) [5pts.] Suppose f : R → R is continuous at 0 and that f(0) 6= 0.Prove that {x ∈ R : f(x) 6= 0} is an infinite set.Help! -δ with a well-chosen  will help you;ideas similar to 1.a) and b) of the homework helped me.Solution: Let  = |f(0)|. Then  > 0.Since f is continuous at 0, we can find a δ > 0 so that|f(x) − f(0)| <  whenever |x − 0| < δ.i.e. |f (x) − f (0)| < |f (0)| whenever |x| < δ. This means(−δ, δ) ⊂ {x ∈ R : f (x) 6= 0}.(b) [5pts.] Suppose that f : [0, ∞) → R is continuous at 0 and thatf(x) ≥ 0 for all x ∈ (0, ∞).Prove that f(0) ≥ 0.You may use any facts that were proved in the homework, as long as you cite themwhen you use them.Help! One sequence really, really helped me out.Solution: Let (xn)∞n=1be the sequence defined by xn=1n.Since f is continuous at 0, each xn≥ 0, and limn→∞xn= 0, we havef(0) = limn→∞f(xn).Since xn> 0 for all n ∈ N, f(xn) ≥ 0 for all n ∈ N,and so problem 1.b) of the homework tells us that f (0) ≥