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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 "$f(x)+g(x)>L−1+M−L+1=M$"

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  • $\begingroup$ The third, fourth and fifth lines were copied from the book verbatim? Dump it. $\endgroup$ – Git Gud Jul 28 '14 at 13:51
  • $\begingroup$ @Git Gud Well, actually the book is not written in English, I tried to translate it the best I could. $\endgroup$ – JaVaPG Jul 28 '14 at 14:00
  • $\begingroup$ @Git Gud Can you explain what not clear? $\endgroup$ – JaVaPG Jul 28 '14 at 14:01
  • $\begingroup$ 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. $\endgroup$ – JaVaPG Jul 28 '14 at 14:19
  • $\begingroup$ I think a proof by cases is implied. I wrote an answer below that hopefully clarifies things a bit. $\endgroup$ – Git Gud Jul 28 '14 at 17:34
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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.

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