Petrus said:
Thanks!
It Also say to proof sum,product and Quotient rule but end with (only the case $$x->+\infty$$ need to be treated) What do they mean with that..?
for $$x->a$$ My explain:
let f be real valued function that is defined in a open interval around a point a $$a \in \mathbb{R}$$ without the point a. Then we say limits for f then x goes against x is equal to L and write as
$$\lim_{x->a}=L$$
for each $$\epsilon > 0 $$ there exist a $$\delta > 0 $$ such that
$$0<|x-a|<\delta => |f(x)-L|< \epsilon$$
notice that $$|x-a|$$ is the distance between the point x and the point a and $$|f(x)-L|$$ is the distance between the number f(x) and L is this correct explain?
Regards,
$$|\pi\rangle$$
For starters, that's how I would define $\displaystyle\lim_{x\to a}f(x)=L$. On the other hand, we have 3 cases to consider for infinite limits:
1. We have $\displaystyle \lim_{x\to a}f(x)=\pm\infty\iff \forall\,E>0,\,\exists\,\delta>0: 0<|x-a|<\delta\implies |f(x)|> E$.
2. We have $\displaystyle \lim_{x\to \pm\infty}f(x)=L \iff \forall\,\epsilon>0,\,\exists\, D>0: |x|>D \implies |f(x)-L|<\epsilon$
3. We have $\displaystyle\lim_{x\to \pm\infty}f(x)=\pm\infty \iff \forall\, E>0,\,\exists\, D>0: |x|>D \implies |f(x)|>E$
With all that said, to me, it seems like you're supposed to use one of these definitions to prove:
a. The sum/difference rule: If $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M$, then $\displaystyle\lim_{x\to a}f(x)\pm g(x) = L\pm M$ (or the equivalent form for $x\to\infty$).
b. The product rule: If $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M$, then $\displaystyle\lim_{x\to a}f(x)g(x) = LM$ (or the equivalent form for $x\to \infty$).
c. The quotient rule: If $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M\neq 0$, then $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M}$ (or the equivalent form for $x\to\infty$).
I'll show product rule in the case of $x\to \infty$. So, we know that
\[\lim_{x\to\infty}f(x)=L\iff \forall\,\epsilon>0,\exists\,D_1>0: x>D_1\implies |f(x)-L|<\epsilon\]
and
\[\lim_{x\to\infty}g(x)=M\iff \forall\,\epsilon>0,\exists\,D_2>0: x>D_2\implies |g(x)-M|<\epsilon\]
(we will come back to this part in a bit because we're going to need to use each of these guys twice.)
Ideally, we would like to see the product $(f(x)-L)(g(x)-M)$ appear in the $|f(x)g(x)-LM|$ term. To get it to appear, we note that $(f(x)-L)(g(x)-M)=f(x)g(x)-Mf(x)-Lg(x) + LM$ and thus
\[\begin{aligned}f(x)g(x)-LM &= (f(x)-L)(g(x)-M) + Mf(x) + Lg(x) - 2LM\\ &= (f(x)-L)(g(x)-M) + Mf(x)-LM + Lg(x) - LM\\ &= (f(x)-L)(g(x)-M) + M(f(x)-L) + L(g(x)-M)\end{aligned}\]
Therefore,
\[|f(x)g(x)-LM| \leq \color{red}{|f(x)-L|}\color{blue}{|g(x)-M|} + |M|\color{green}{|f(x)-L|} + |L|\color{purple}{|g(x)-M|}.\]
Now, we use the facts that $\displaystyle\lim_{x\to\infty}f(x)=L$ and $\displaystyle\lim_{x\to \infty}g(x)=M$ to see that
\[\forall\,\epsilon>0,\,\exists \color{red}{D_1}>0: x>D_1 \implies \color{red}{|f(x)-L|}<\color{red}{\sqrt{\frac{\epsilon}{3}}}\]
\[\forall\,\epsilon>0,\,\exists \color{blue}{D_2}>0: x>D_2 \implies \color{blue}{|g(x)-M|}<\color{blue}{\sqrt{\frac{\epsilon}{3}}}\]
\[\forall\,\epsilon>0,\,\exists \color{green}{D_3}>0: x>D_3 \implies \color{green}{|f(x)-L|}<\color{green}{\frac{\epsilon}{3|M|}}\]
\[\forall\,\epsilon>0,\,\exists \color{purple}{D_4}>0: x>D_4 \implies \color{purple}{|g(x)-M|}<\color{purple}{\frac{\epsilon}{3|L|}}\]
Therefore, if we take $D=\max\{D_1,D_2,D_3,D_4\}$, we now see that
\[\forall\,\epsilon>0,\,\exists\,D>0:x>D\implies |f(x)g(x)-LM| < \color{red}{\sqrt{\frac{\epsilon}{3}}} \color{blue}{\sqrt{\frac{\epsilon}{3}}} + |M|\color{green}{\frac{\epsilon}{3|M|}} + |L|\color{purple}{\frac{\epsilon}{3|L|}} = \frac{\epsilon}{3}+ \frac{\epsilon}{3}+ \frac{\epsilon}{3} = \epsilon.\]
Hence, $\displaystyle\lim_{x\to\infty} f(x)g(x) = \lim_{x\to\infty}f(x)\cdot\lim_{x\to\infty}g(x) = LM$.
I hope this makes sense! (Sun)
