MATH 0420 Midterm: e1(0420)_2014_spring_sln
Math 0420 Midterm Exam
Spring 2014 S o l u t i o n s
1. Let S⊂Rand x∈R. Assume that xis a cluster point of S. Show that there exists a convergent
sequence {xn} ⊂ Ssuch that xn6=xand lim xn=x.
Solution: See Proposition 3.1.2.
2. Let f: [1,3] →Rbe defined by f(x) = 1
x.
(a) Using ε-δdefinition of a limit show that lim
x→2f(x) = 1
2.
Solution:
1
x−1
2
=|x−2|
2x. 1 ≤x≤3, 2 ≤2x≤6, 1
6≤1
2x≤1
2.
Then ∀ε > 0∃δ= 2εs.t. |x−2|< δ ⇒
1
x−1
2
=|x−2|
2x≤|x−2|
2<δ
2=2ε
2=ε
(b) Is f(x) continuous at 2 ? Support your answer.
Solution: f(2) = 1
2. We have lim
x→2f(x) = f(2) and hence fis continuous at 2.
3. Let f:S→Rbe a uniformly continuous function and {xn} ⊂ Sbe a Cauchy sequence.
Is the sequence {f(xn)}Cauchy? Support your answer.
Solution: See Lemma 3.4.5.
4. (a) (10 points) Is the function f(x) = x|x|differentiable at 0?
Support your answer by using the definition of differentiability of a function.
Solution: lim
x→0
f(x)−f(0)
x−0= lim
x→0
x|x|
x= lim
x→0|x|= 0.
The limit of the difference quotient exists. Therefore the function is differentiable at 0.
(b) (10 points) Is the function f(x) = |x|sin xdifferentiable at −1? Support your answer.
You may use any method.
Document Summary
[1, 3] r be de ned by f (x) = Using - de nition of a limit show that lim x 2 f (x) = 1 x 3, 2 2x 6, Then > 0 = 2 s. t. We have lim x 2 f (x) = f (2) and hence f is continuous at 2: let f : s r be a uniformly continuous function and {xn} s be a cauchy sequence. Support your answer by using the de nition of di erentiability of a function. Solution: lim x 0 f (x) f (0) x 0. Therefore the function is di erentiable at 0. (b) (10 points) |x| = x when x is near 1. Then f (x) = x sin x is di erentiable as a product: let f (x) = x7 + 3x3 + x 1. (a) Show that f (x) is strictly increasing, and therefore invertible.
