If f(x) -> oo and g(x) -> c, prove that f(x)+g(x) -> oo

[issacnewton]

Homework Statement

If ##\lim_{x \rightarrow a} f(x) = \infty ## and ## \lim_{x \rightarrow a} g(x) = c ##, then prove that
$$ \lim_{x \rightarrow a} f(x) +g(x) = \infty $$

Homework Equations

Epsilon Delta definition of limit

The Attempt at a Solution

I will take two cases here.
Case 1. ## c \geqslant 0##
Let ##M>0## be arbitrary. Then ## 2M > 0##. Since we have ##\lim_{x \rightarrow a} f(x) = \infty ## , ##\exists~\delta_1 >0 ## such that
$$ \forall~x \in D(f) \left[ 0 < |x-a| < \delta_1 \rightarrow f(x) > 2M \right] \cdots\cdots (1)$$
Where, ##D(f)## is the domain of function ##f##. We also have ##\frac M 2 > 0##. Since ## \lim_{x \rightarrow a} g(x) = c ##, ##\exists~\delta_2 >0 ## such that
$$ \forall~x \in D(g) \left[ 0 < |x-a| < \delta_2 \rightarrow |g(x)-c| < \frac M 2 \right] \cdots\cdots (2)$$
where ##D(g)## is the domain of function ##g##. Now let ##\delta = \text{min}(\delta_1, \delta_2)##. Let ##x_1 \in D(f+g)## be arbitrary. And also suppose that ## 0 < |x-a| < \delta ##. Since ##x_1 \in D(f+g)##, we have that ## x_1 \in D(f) ## and ## x_1 \in D(g) ##. Also since ## 0 < |x-a| < \delta ##, from the definition of
##\delta##, it implies that ## 0 < |x-a| < \delta_1## and ## 0 < |x-a| < \delta_2 ##. So we can conclude from the equation 1 and 2 that, ## f(x_1) > 2M## and ## |g(x_1) -c| < \frac M 2##. It follows that $$ c - \frac M 2 < g(x_1) < c + \frac M 2$$ $$\therefore f(x_1) + c - \frac M 2 <f(x_1)+ g(x_1) < f(x_1) + c + \frac M 2 \cdots\cdots(3)$$ But we have ##f(x_1) > 2M##, so ##f(x_1) - \frac M 2 > \frac{3M}{2}##. This leads to ## f(x_1) - \frac M 2 + c > \frac{3M}{2} + c##. Since ## c \geqslant 0##, we get ## \frac{3M}{2} + c \geqslant \frac{3M}{2} > M##. We conclude that ## f(x_1) - \frac M 2 + c > M##. Using this in equation ##(3)##, we have $$ M < f(x_1) + c - \frac M 2 < f(x_1)+ g(x_1)$$ So we conclude that ## f(x_1)+ g(x_1) > M ##. Since ##M## and ##x_1## are arbitrary, we conclude that $$ \lim_{x \rightarrow a} f(x) +g(x) = \infty $$ Now we proceed to the case 2.
Case 2. ## c<0##
Again let ##M > 0## be arbitrary. So we have ##-2c >0## and ##2M >0##. So ##2M - 2c >0##. Since we have ##\lim_{x \rightarrow a} f(x) = \infty ## , ##\exists~\delta_1 >0 ## such that $$ \forall~x \in D(f) \left[ 0 < |x-a| < \delta_1 \rightarrow f(x) > (2M-2c) \right] \cdots\cdots (4)$$ Where, ##D(f)## is the domain of function ##f##. Since ##-c>0##, and since ## \lim_{x \rightarrow a} g(x) = c ##, ##\exists~\delta_2 >0 ## such that $$ \forall~x \in D(g) \left[ 0 < |x-a| < \delta_2 \rightarrow |g(x)-c| < -c \right] \cdots\cdots (5)$$ where ##D(g)## is the domain of function ##g##. Now let ##\delta = \text{min}(\delta_1, \delta_2)##. Let ##x_1 \in D(f+g)## be arbitrary. And also suppose that ## 0 < |x-a| < \delta ##. Since ##x_1 \in D(f+g)##, we have that ## x_1 \in D(f) ## and ## x_1 \in D(g) ##. Also since ## 0 < |x-a| < \delta ##, from the definition of ##\delta##, it implies that ## 0 < |x-a| < \delta_1## and ## 0 < |x-a| < \delta_2 ##. So we can conclude from the equation 4 and 5 that, ## f(x_1) > 2M-2c ## and ## |g(x_1) -c| < -c ##. So we get ## c < g(x_1) - c < -c##. Which is ## 2c < g(x_1) < 0 ##. Therefore, ## f(x_1) + 2c < f(x_1) + g(x_1)##. Since## f(x_1) > 2M-2c ##, we have $$ M< 2M < f(x_1)+2c < f(x_1) + g(x_1)$$ From here we conclude that ## f(x_1) + g(x_1) > M##. Since ##M## and ##x_1## are arbitrary, we conclude that $$ \lim_{x \rightarrow a} f(x) +g(x) = \infty $$ So we have proven this for both cases and hence ##\forall ~c \in \mathbb{R}##, we can say that $$ \lim_{x \rightarrow a} f(x) +g(x) = \infty $$ Can you check if this is OK!
Thanks

 

[Delta2]

Seems correct to me but I don't see the reason you introduce ##x_1##, you should just say that for every x such that ##0<|x-a|<\delta## and proceed with x instead of x1.

 

[issacnewton]

Well Delta, Since we have a universal quantifier in the limit definition, I choose an arbitrary ##x_1## and then proceed to prove the statement inside that quantifier. After we prov the statement , then we can say that the statement is true for arbitrary ##x_1##, it is true for every x in the domain

 

[Delta2]

Ok then I see a little mistake you say ##x_1## arbitrary but it isn't completely arbitrary it is such that ##0<|x_1-a|<\delta## so that (1) (2) or (4) and (5) are true for x1. You are saying this but you using x instead of x1.

 

[issacnewton]

Delta , I think the original goal here is to prove that $$\forall ~M >0 \exists~ \delta >0~ \forall ~x \in D(f+g)\left[ 0<|x-a| < \delta \rightarrow f(x)+g(x) > M \right] $$ So initially, I let ##M > 0## be arbitrary. Then with the choice of ##\delta## I have done, my goal becomes, $$\forall ~x \in D(f+g)\left[ 0<|x-a| < \delta \rightarrow f(x)+g(x) > M \right] $$ Since there is a universal quantifier at the beginiing, I let ## x \in D(f+g)## as arbitrary. So now the goal becomes $$0<|x-a| < \delta \rightarrow f(x)+g(x) > M $$ Since this is an implication, I assume the antecedent, ## 0<|x-a| < \delta## and then proceed to prove the consequent, which is ## f(x)+g(x) > M##.

 

[Delta2]

Fine it is just that you assume the antecedent for ##x## while you prove the consequent for ##x_1##.

 

[issacnewton]

Actually now I see your point. I have used ##x_1## when I chose an arbitrary member in the domain , but I used ##x## in the antecedent. I should have changed that to ##x_1##. So with that error removed proof seems ok !