# Prove Limit Exist For 3 Statements

Discussion in 'Calculus' started by nycmathguy, May 12, 2022.

1. ### nycmathguy

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Calculus
Section 2.4

We end Section 2.4 with this problem. nycmathguy, May 12, 2022

2. ### MathLover1

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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=-∞

MathLover1, May 12, 2022
nycmathguy likes this.

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