# Calculus - The proof of arithmetic infinity limits

I having trouble to understand the proof of arithmetic infinity limits.

(I'm quoting from my learning book)

f,g are functions and lets assume that :

$$\lim_{x \to x_0}f(x)=L \mbox{ (final)}$$ $$\lim_{x \to x_0} g(x)=\infty$$

Prove that :

$$\lim_{x \to x_0}(f+g)(x)=\infty$$

f,g are defined in $N*\delta(x_0)$ (pocked environment)

We need to show that for all M>0 exist $\delta_*$>0 so all $x$ that appiles $0<|x-x_0|<\delta_*$ appiles $(f+g)(x)>M$

there is M>0 big enough that $M>L-1$.

$$\lim_{x \to x_0}g(x)=\infty$$ Therefore, exist $0<\delta_1<\delta,\delta_1$ so all x that appiles $0<|x-x_0|<\delta_1$ appiles $g(x)>M-L+1$

As well, $$\lim_{x \to x_0}f(x)=L$$

Therefore exist $0<\delta_2<\delta,\delta_2$ so all x appiles $0<|x-x_0|<\delta_2$ appiles $|f(x)-L|<1$ so $f(x)>L-1 We choose$\delta_*$=min{$\delta_1,\delta_2$} therefore for all x appiles$0<|x-x_0|<\delta_*$appiles :$(f+g)(x) = f(x)+g(x) > L-1+M-L+1=M$. I don't understand how there is$M>0$that$M>L-1$? In addition that statement "$f(x)+g(x)>L−1+M−L+1=M$" • The third, fourth and fifth lines were copied from the book verbatim? Dump it. – Git Gud Jul 28 '14 at 13:51 • @Git Gud Well, actually the book is not written in English, I tried to translate it the best I could. – JaVaPG Jul 28 '14 at 14:00 • @Git Gud Can you explain what not clear? – JaVaPG Jul 28 '14 at 14:01 • I'll be more clearly, by 'final' I mean that the limit isn't infinite so your interpretation is correct$L\in R$. this $$\lim_{x \to x_0}(f+g)(x)=\infty$$ is what should be proven. – JaVaPG Jul 28 '14 at 14:19 • I think a proof by cases is implied. I wrote an answer below that hopefully clarifies things a bit. – Git Gud Jul 28 '14 at 17:34 ## 1 Answer The statement to prove is the following. Let$I$be a non-empty subset of the real numbers which contains an interval of the form$[a,+\infty[$, for some$a\in \mathbb R$,$x_0\in \overline I$and$f,g\colon I\to \mathbb R$functions such that$\lim \limits_{x\to x_0}\left(g(x)\right)=+\infty$and$\lim \limits_{x\to x_0}\left(f(x)\right)=L$, for some$L\in \mathbb R$. In these conditions it holds true that$\lim \limits_{x\to x_0}\left((f+g)(x)=+\infty\right)$. To prove this recall that $$\lim \limits_{x\to x_0}\left(f(x)\right)=L\iff \forall \varepsilon >0\,\exists \delta _2>0\,\forall x\in I\left(0<|x-x_0|<\delta _2\implies |f(x)-L|<\varepsilon\right)$$ and $$\lim \limits_{x\to x_0}(g(x))=+\infty\iff \forall \color{purple}M>0\,\exists \delta _1>0\,\forall x\in I(0<|x-x_0|<\delta _1\implies g(x)>\color{purple}M).$$ Remember the goal is to prove that$\lim \limits_{x\to x_0}\left((f+g)(x)=+\infty\right)$or equivalently $$\forall \color{blue}M>0\,\exists \delta_*>0\,\forall x\in I(0<|x-x_0|<\delta _*\implies (f+g)(x)>\color{blue}M).$$ Proof: Begin by taking an arbitrary (blue)$\color{blue}M>0$. Either$\color{blue}M\leq L-1$holds or$\color{blue}M>L-1$does.$\bbox[5px,border:2px solid #000000]{\text{Case: }\color{blue}M> L-1}$The goal is to find$\delta _*>0$such that$\forall x\in I(0<|x-x_0|<\delta _*\implies g(x)>\color{blue}M)$. With$\varepsilon=1$one gets the existence of$\delta _2>0$with the property that$\forall x\in I(0<|x-x_0|<\delta _2\implies |f(x)-L|<1)$. With$\color{purple}M=\color{blue}M-(L-1)\color{grey}{>0}$. one gets the existence of$\delta _1>0$such that$\forall x\in I(0<|x-x_0|<\delta _1\implies g(x)>\color{blue}M-(L-1))$. Now define$\delta _*:=\min\left(\{\delta _1, \delta _2\}\right)$. The goal is now to prove that$\forall x\in I(0<|x-x_0|<\delta _*\implies (f+g)(x)>\color{blue}M)$. Take$x\in I$and assume that$0<|x-x_0|<\delta _*$. Since$\delta _*\leq \delta _2$one gets$|f(x)-L|<1$, i.e.,$-1<f(x)-L<1$which implies$L-1<f(x)$. Since$\delta _*\leq \delta _1$one gets$\color{blue}M-(L-1)<g(x)$. Therefore$L-1+\color{blue}M-(L-1)<f(x)+g(x)$, that is,$(f+g)(x)>\color{blue}M$.$\bbox[5px,border:2px solid #000000]{\text{Case: }\color{blue}M\leq L-1}\$

Hopefully, after reading the case above, you can get an idea to solve this case.