Prove Limit Exist For 3 Statements

[nycmathguy]

Calculus
Section 2.4

We end Section 2.4 with this problem.

 

[MathLover1]

44.

suppose that lim(f(x))=∞ as x->a and lim(g(x)=c as x->a, where c is real number. prove each statement

a. lim[f(x) +g(x)]=∞ as x->a

Let M element of R. Then, we need to show that:

all δ element of R: 0<|x−c|<δ => f(x)+g(x) >M

Since lim(f(x))=∞ as x→a, it follows that:

0<|x−c|<δ
|(f(x) -L)<L/2

So, that means that:

0<|x−c|<δ2 =>|f(x) |>L/2

Define δ=min{δ1,δ2}. Then:

0<|x−c|<δ =>f(x)+g(x)>2M/L*L/2
0<|x−c|<δ =>f(x)+g(x)>M which proves that
lim(f(x) +g(x))=∞as x->a

so, lim[f(x) +g(x)]=∞ as x->a =infinity +c=∞


b.
lim[f(x) *g(x)]=∞ as x->a if c>0

lim(f(x)*g(x), x→a
=lim[[f(x)−K][g(x)−L]+Lf(x)+Kg(x)−KL], x→a
=lim[f(x)−K][g(x)−L], x→a +lim(Lf(x)),x→a +lim(Kg(x)),x→a −lim(KL), x→a
=0+lim(Lf(x)),x→a+lim(Kg(x)),x→a−lim(KL),x→a
=LK+KL−KL
=LK

in your case L=∞ and K=c, => LK=∞*c=∞

so, lim[f(x) *g(x)]=∞ as x->a if c>0

c.
lim[f(x) *g(x)]= -∞ as x->a if c<0

all same, just c<0
in your case L=∞ and K=-c, => LK=∞*-c=-∞

 

[nycmathguy]

44.

suppose that lim(f(x))=∞ as x->a and lim(g(x)=c as x->a, where c is real number. prove each statement

a. lim[f(x) +g(x)]=∞ as x->a

Let M element of R. Then, we need to show that:

all δ element of R: 0<|x−c|<δ => f(x)+g(x) >M

Since lim(f(x))=∞ as x→a, it follows that:

0<|x−c|<δ
|(f(x) -L)<L/2

So, that means that:

0<|x−c|<δ2 =>|f(x) |>L/2

Define δ=min{δ1,δ2}. Then:

0<|x−c|<δ =>f(x)+g(x)>2M/L*L/2
0<|x−c|<δ =>f(x)+g(x)>M which proves that
lim(f(x) +g(x))=∞as x->a

so, lim[f(x) +g(x)]=∞ as x->a =infinity +c=∞


b.
lim[f(x) *g(x)]=∞ as x->a if c>0

lim(f(x)*g(x), x→a
=lim[[f(x)−K][g(x)−L]+Lf(x)+Kg(x)−KL], x→a
=lim[f(x)−K][g(x)−L], x→a +lim(Lf(x)),x→a +lim(Kg(x)),x→a −lim(KL), x→a
=0+lim(Lf(x)),x→a+lim(Kg(x)),x→a−lim(KL),x→a
=LK+KL−KL
=LK

in your case L=∞ and K=c, => LK=∞*c=∞

so, lim[f(x) *g(x)]=∞ as x->a if c>0

c.
lim[f(x) *g(x)]= -∞ as x->a if c<0

all same, just c<0
in your case L=∞ and K=-c, => LK=∞*-c=-∞

Great as usual. Keep in mind that the problems posted are those I didn't quite understand in the textbook. I also don't have sufficient time during the week to show my math (right or wrong) as it may be. It's not laziness. Trust me, the overnight shift is not for everybody.