Math Problem Solver

Recall that Proposition 3.1 from the lecture notes states that if

∙

real numbers a,L∈R

and

∙

a function f

that exists on a punctured open interval about the point a

are such that there are a positive number δ0>0

and a positive constant C>0

satisfying

0<|x−a|<δ0⇒|f(x)−L|<C|x−a|(1.1)

for all x∈dom(f),

then for every ε>0,

0<|x−a|<min(δ0,εC)⇒|f(x)−L|<ε

for all x∈dom(f).

In particular, we have that

limx→af(x)=L,

since for every ε>0,

a required δ=δ(ε)>0

from the definition of the limit of a function at a given point can be chosen as

min(δ0,εC).

As it has been discussed during the lectures, one may try to use one of the values

δ0=1,12,14,…

to see whether a necessary constant C

satisfying (1.1) can be found.

(Limit of a Function, the Definition of). (i) Fill in the blanks in the following argument that shows that

limx→4x2=16.

(a) Let f(x)=x2

and let L=16.

Observe that

|f(x)−L|=|x2−16|=|(x−4)(x+4)|=|x+4|⋅|x−4|.

(b) Keeping in mind application of Proposition 3.1, we then wish to obtain the least upper bound C

of the function

g(x)=|x+4|

on

(3,5)={x:|x−4|<1}.

For every x∈R,

we have that

|x−4|<1⇒3<x<5⇒x∈(3,5)⇒g(x)=|x+4|=x+4<C,

where

C=

Answer 1 Question 1

(please find and then enter the least positive number which satisfies the last inequality for all x∈(3,5)

).

Accordingly,

|x−4|<1⇒|f(x)−L|<C|x−4|

for all x∈R.

(c) Thus we see that Proposition 3.1 is applicable for

δ0=

Answer 2 Question 1

and C=

Answer 3 Question 1

,

and then function f(x)=x2

does indeed converge to L=16

as x

approaches 4,

as we wished to prove.

(ii) Fill in the blanks in the following argument that shows that

limx→8x−6x−3=25.

(a) Let

f(x)=x−6x−3

and L=2/5.

Observe that

|f(x)−L|=∣∣∣x−6x−3−25∣∣∣=q|x−3|⋅|x−8|

for all x≠3,

where the positive rational number q

is given by

q=

Answer 4 Question 1

.

(b) Keeping in mind application of Proposition 3.1, we then wish to obtain the least upper bound C

of the function

g(x)=q|x−3|

on

(7,9)={x:|x−8|<1}.

For every x∈R,

we have that

|x−8|<1⇒7<x<9⇒x∈(7,9)⇒g(x)=q|x−3|<C,

where

C=

Answer 5 Question 1

(please find and then enter the least positive rational number which satisfies the last inequality for all x∈(7,9)

).

Accordingly,

|x−8|<1⇒|f(x)−L|<C|x−8|

for all x∈R.

(c) Thus we see that Proposition 3.1 is applicable for

δ0=

Answer 6 Question 1

and C=

Answer 7 Question 1

,

and then function

f(x)=x−6x−3

does indeed converge to L=2/5

as x

approaches 8,

as we wished to prove.