Question

prove that if:

Answer 1

What do we need to prove?

Answer 2

\[\lim_{x \rightarrow a} f(x)= l\] and \[\lim_{x \rightarrow a} g(x)= m\] then,
\[\lim_{x \rightarrow a}(f+g)(x)=l+m\]
\[\lim_{x \rightarrow a}(fg)(x)=lm\]
moreover,
if \[m \neq 0\]
then,
\[\lim_{x \rightarrow a}(\frac{ 1 }{ g })(x)=\frac{ 1 }{ m }\]

Answer 3

sorry that it took a while to post, I am in the middle of work as well >.<

Answer 4

Presumably this would be a proof using \(\epsilon-\delta\) definitions?

Answer 5

you know your stuff, KG
LOL

Answer 6

Assuming that is the case, we know that\[\forall\epsilon>0,\exists\delta_1>0 \text{ s.t. } \forall x\neq a,\,|x-a|<\delta_1\implies|f(x)-l|<\epsilon,\]and\[\forall\epsilon>0,\exists\delta_2>0 \text{ s.t. } \forall x\neq a,\,|x-a|<\delta_2\implies|g(x)-m|<\epsilon.\]Conider the function \((f+g)(x)\). Then consider \[|(f+g)(x)-(l+m)|=|f(x)-l+g(x)-m|.\]By the triangle inequality, this is less than or equal to \[|f(x)-l|+|g(x)-m|.\]Thus, if \(|x-a|<\max\{\delta_1,\delta_2\}\), then \(|(f+g)(x)-(l+m)|<2\epsilon\), so \(\lim_{x\to a}(f+g)(x)=l+m\).

Answer 7

If you're unhappy with using \(2\epsilon\), you can redefine \(\epsilon\) to be \(\epsilon/2\).

Answer 8

Before you read the next part, there's a minor correction above. Instead of \(\max\{\delta_1,\delta_2\}\), we should be using the minimum instead.

If we now move onto \((fg)(x)\), then we must consider \(|(fg)(x)-lm|=|f(x)g(x)-lm|\). But we're going to play a little trick, where instead look at the limits of \(f(x)-l\) and \(g(x)-m\). We know that\[\forall\epsilon>0,\exists\delta_1>0 \text{ s.t. } \forall x\neq a,\,|x-a|<\delta_1\implies|(f(x)-l)-0|<\epsilon\]and that\[\forall\epsilon>0,\exists\delta_2>0 \text{ s.t. } \forall x\neq a,\,|x-a|<\delta_2\implies|(g(x)-m)-0|<\epsilon.\]I.e. that
\[\lim_{x\to a}f(x)-l=0\]and that\[\lim_{x\to a}g(x)-m=0\]Now consider \(|f(x)-l||g(x)-m|\). By the above two limits, we know that this is less than or equal to \(\epsilon^2\) when \(|x-a|<\min\{\delta_1,\delta_2\}\).

Answer 9

Next, we write \(f(x)g(x)=(f(x)-l)(g(x)-m)+lg(x)+mf(x)-lm\), and take the limit.\[\begin{aligned}
\lim_{x\to a}(f(x)-l)& (g(x)-m)+lg(x)+mf(x)-lm\\&=\lim_{x\to a}(f(x)-l)(g(x)-m)+\lim_{x\to a}lg(x)+\lim_{x\to a}mf(x)-\lim_{x\to a}lm\\
&=0+lm+ml-lm\\
&=lm.
\end{aligned}\]

Answer 10

Onto the last part. We know that\[\forall\epsilon>0,\exists\delta_1>0 \text{ s.t. } \forall x\neq a,\,|x-a|<\delta_1\implies|g(x)-m|<\epsilon\]and that \(m\neq 0\). Then we can say that \(|g(x)-m|+|g(x)|\ge |m|\) since \(|g(x)-m|=|m-g(x)|\), and from the triangle inequality. So using the fact that \(|g(x)-m|<\epsilon\), this tells us that\[|m|<|g(x)|+\epsilon\implies\frac{1}{|g(x)|}<\frac{\epsilon}{|m|}.\]Consider\[\left\vert\frac{1}{g(x)}-\frac{1}{m}\right\vert=\left\vert\frac{m-g(x)}{mg(x)}\right\vert=\frac{1}{|mg(x)|}|m-g(x)|=\frac{1}{|m|}\frac{1}{|g(x)|}|m-g(x)|.\]From what we just showed, this is strictly less than\[\frac{1}{|m|}\frac{\epsilon}{|m|}|m-g(x)|<\frac{\epsilon}{|m|^2}\epsilon=\frac{\epsilon^2}{|m|^2}.\]Since \(m\) is fixed and non zero, this means that \[\lim_{x\to a}\frac{1}{g(x)}=\frac{1}{m}.\]Again, if you're unhappy with having the \(\epsilon^2\), we can just redefine \(\epsilon=\frac{\epsilon}{|m|}\) at the start.

Answer 11

thank you.
I will look at this in a little bit.

Answer 12

Just fyi, I did end up having to reference http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx for the last limit. He chooses values in such a way that he ends with an \(\epsilon\) instead of what I got.