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 "$f(x)+g(x)>L−1+M−L+1=M$"

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.