MATH 0420 Midterm: q1(0420)_2014_spring_sln
Math 0420 Quiz 1
Spring 2014 S o l u t i o n s
1. Find an example of functions fand gsuch that both limits lim
x→1f(x) and lim
x→1g(x) do not exist
but lim
x→1f(x)g(x) exists.
Solution: For example,
f(x) = (x, if x6= 1
2,if x= 1 g(x) = (x, if x6= 1
1
2,if x= 1 then f(x)g(x) = (x2,if x6= 1
1,if x= 1 =x2
2. Let f:R→Rbe continuous. Suppose that f(c)>0 for some c∈R. Show that there exists an
a > 0 such that for all x∈(c−a, c +a) we have f(x)>0.
Solution: fis continuous at c, i.e. ∀ε > 0∃δ > 0 s.t. if |x−c|< δ then |f(x)−f(c)|< ε. The
last inequality is equivalent to the double inequality f(c)−ε < f(x)< f(c) + ε.
Take ε=f(c)>0. Then ∃a=δ > 0 s.t. if |x−c|< a then f(c)−f(c)< f(x)< f(c) + f(c).
The last means, if x∈(c−a, c +a) then we have f(x)>0.
Document Summary
S o l u t i o n s: find an example of functions f and g such that both limits lim x 1 f (x) and lim x 1 g(x) do not exist but lim x 1 f (x)g(x) exists. 2, if x 6= 1 if x = 1 g(x) =(x, 1 if x 6= 1 if x = 1 then f (x)g(x) =(x2, = x2: let f : r r be continuous. Suppose that f (c) > 0 for some c r. show that there exists an a > 0 such that for all x (c a, c + a) we have f (x) > 0. Then a = > 0 s. t. if |x c| < a then f (c) f (c) < f (x) < f (c) + f (c).
