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 !