MATH 1300 Study Guide - Midterm Guide: Intermediate Value Theorem, Squeeze Theorem
1 Oct 2018
School
Department
Course
Professor
Mathematics 1300 3.0 AF
Solutions to Test 1
1. Solution: (i) Let x→0 along the numbers 1
n+1
2π. Then
sin 1
x= sin n+1
2π= sin nπ cos π
2+ cos nπ sin π
2= (−1)n.
So, limx→0sin 1
xdoes not exist.
Solution: (ii) Since
−x2≤x2sin 1
x≤x2
and limx→0x2= 0,it follows from the squeeze theorem that
lim
x→0x2sin 1
x= 0.
2. Solution: |f(x)−8|<0.001 if |3x+ 2 −8|<0.001 or if |3x−6|<0.001
or if 3|x−2|<0.001 or if |x−2|<0.001/3. So, let δbe any positive
number ≤0.001/3. Then
0<|x−2|< δ ⇒ |f(x)−8|= 3|x−2|<3δ= 0.001.
3. Solution: Let f(x) = ex−3+2x. Then f(0) = −2 and f(1) = e−3 +
2+1.7.Therefore 0 is between f(0) and f(1) By the Intermediate Value
Theorem, there exists a number cin (0,1) such that
f(c) = 0.
Therefore
ec= 3 −2c.
4. Solution:
f0(x) = lim
h→0
f(x+h)−f(x)
h
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MATH 1300 Full Course Notes
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Document Summary
Solutions to test 1: solution: (i) let x 0 along the numbers (cid:19) = sin n cos. So, limx 0 sin 1 x does not exist. 1 x and limx 0 x2 = 0, it follows from the squeeze theorem that x2sin lim x 0. = 0: solution: |f (x) 8| < 0. 001 if |3x + 2 8| < 0. 001 or if |3x 6| < 0. 001 or if 3|x 2| < 0. 001 or if |x 2| < 0. 001/3. So, let be any positive number 0. 001/3. 0 < |x 2| < |f (x) 8| = 3|x 2| < 3 = 0. 001: solution: let f (x) = ex 3 + 2x. Then f (0) = 2 and f (1) = e 3 + Therefore 0 is between f (0) and f (1) by the intermediate value. Theorem, there exists a number c in (0, 1) such that.
