# Prove limit theorem using epsilon-delta

## Main Question or Discussion Point

Hey i am trying to understand Spivak's proof of lim x->a of f(x)g(x)=lm (where l is limit of f(x) and m is lim of g(x) )..but i think he is skipping many steps and at one point i dont understand why he is doing something..

ok so the following i understand:

$\left|f(x)g(x)-lm\right|$< E
$\leq$$\left|f(x)\right|$$\left|g(x)-m\right|$ + $\left|m\right|$$\left|f(x)-l\right|$<E

$\left|f(x)-l\right|$$\leq$1
$\left|f(x)\right|$$\leq$$\left|l\right|$+1

$\left|f(x)\right|$$\left|g(x)-m\right|$<($\left|l\right|$+1)$\left|g(x)-m\right|$<E/2
$\left|g(x)-m\right|$< $\frac{E}{2(\left|l\right|+1)}$

$\left|f(x)\right|$$\left|g(x)-m\right|$<($\left|l\right|$+1)$\frac{E}{2(\left|l\right|+1)}$

$\left|f(x)\right|$$\left|g(x)-m\right|$<E/2

ok so now i am guessing that we wanna say that $\left|m\right|$$\left|f(x)-l\right|$<E/2 so that E/2 + E/2 = E
so can u just say:
$\left|m\right|$$\left|f(x)-l\right|$<E/2
$\left|f(x)-l\right|$<E/(2$\left|m\right|$)

cus i mean m is just a constant..it cant be restricted like f(x) was...

so:
$\left|m\right|$$\left|f(x)-l\right|$<$\left|m\right|$(E/(2$\left|m\right|$))= E/2

so then:
$\left|f(x)\right|$$\left|g(x)-m\right|$ + $\left|m\right|$$\left|f(x)-l\right|$< E/2 + E/2 =E

when $\left|f(x)-l\right|$ < min (1, E/(2|m|) ) and $\left|g(x)-m\right|$<$\frac{E}{2(\left|l\right|+1)}$

but Spivak is saying that $\left|f(x)-l\right|$ < min (1, E/( 2(|m|+1) ) ) and i have no clue why... help plsss?

Hey i am trying to understand Spivak's proof of lim x->a of f(x)g(x)=lm (where l is limit of f(x) and m is lim of g(x) )..but i think he is skipping many steps and at one point i dont understand why he is doing something..

ok so the following i understand:

$\left|f(x)g(x)-lm\right|$< E
$\leq$$\left|f(x)\right|$$\left|g(x)-m\right|$ + $\left|m\right|$$\left|f(x)-l\right|$<E

$\left|f(x)-l\right|$$\leq$1
$\left|f(x)\right|$$\leq$$\left|l\right|$+1

$\left|f(x)\right|$$\left|g(x)-m\right|$<($\left|l\right|$+1)$\left|g(x)-m\right|$<E/2
$\left|g(x)-m\right|$< $\frac{E}{2(\left|l\right|+1)}$

$\left|f(x)\right|$$\left|g(x)-m\right|$<($\left|l\right|$+1)$\frac{E}{2(\left|l\right|+1)}$

$\left|f(x)\right|$$\left|g(x)-m\right|$<E/2

ok so now i am guessing that we wanna say that $\left|m\right|$$\left|f(x)-l\right|$<E/2 so that E/2 + E/2 = E
so can u just say:
$\left|m\right|$$\left|f(x)-l\right|$<E/2
$\left|f(x)-l\right|$<E/(2$\left|m\right|$)

cus i mean m is just a constant..it cant be restricted like f(x) was...

so:
$\left|m\right|$$\left|f(x)-l\right|$<$\left|m\right|$(E/(2$\left|m\right|$))= E/2

so then:
$\left|f(x)\right|$$\left|g(x)-m\right|$ + $\left|m\right|$$\left|f(x)-l\right|$< E/2 + E/2 =E

when $\left|f(x)-l\right|$ < min (1, E/(2|m|) ) and $\left|g(x)-m\right|$<$\frac{E}{2(\left|l\right|+1)}$
Yes, but now you have done something like

$$|f(x)-l|<\delta_1~\vert~|g(x)-m|<\delta_2$$

so you have two delta's. But you only want one delta. So if we take $\delta=\min(\delta_1,\delta_2)$, then $|f(x)-l|<\delta$ implies that $|f(x)-l|<\delta_1$. And the same with g(x).

So Spivak has two delta's:

$$\delta=1=\min(1,\varepsilon/(2|m|))~\text{and}~\delta_2=1/(2(|l|+1))$$

and he combines it to one delta. This is what he does.

Yes, but now you have done something like

$$|f(x)-l|<\delta_1~\vert~|g(x)-m|<\delta_2$$

so you have two delta's. But you only want one delta. So if we take $\delta=\min(\delta_1,\delta_2)$, then $|f(x)-l|<\delta$ implies that $|f(x)-l|<\delta_1$. And the same with g(x).

So Spivak has two delta's:

$$\delta=1=\min(1,\varepsilon/(2|m|))~\text{and}~\delta_2=1/(2(|l|+1))$$

and he combines it to one delta. This is what he does.
I know that you have to combine the two deltas by taking the minimum delta, but my question is, how does Spivak get that $\left|f(x)-l\right|$ < min (1, E/( 2(|m|+1) ) ) ?

He doesnt get $\left|f(x)-l\right|$ < min (1, E/( 2(|m|) )

He is adding some random 1 next to the |m| ?

He's combining the delta's

$$\delta_1=\min(1,\frac{1}{2|m|})~\text{and}~\delta_2=\frac{1}{2|m|+1}$$

into

$$\delta=\min(\delta_1,\delta_2)=\min(1,\frac{1}{2|m|+1})$$

He's combining the delta's

$$\delta_1=\min(1,\frac{1}{2|m|})~\text{and}~\delta_2=\frac{1}{2|m|+1}$$

into

$$\delta=\min(\delta_1,\delta_2)=\min(1,\frac{1}{2|m|+1})$$
But delta#2 is not $\frac{1}{2|m|+1}$

it is $\frac{1}{2(|l|+1)}$

Ah yes, it seems you're right then. I suspect Spivak adds the +1 to make sure that he didn't divide by 0. That is: if l=0, then you would have 1/0 if you didn't add +1.

Ah yes, it seems you're right then. I suspect Spivak adds the +1 to make sure that he didn't divide by 0. That is: if l=0, then you would have 1/0 if you didn't add +1.
haha its legal to just do that? i mean this is supposed to be a rigid proof...i dont think u can just add a random 1 without explaining why?

why not just say that m$\neq0$, l$\neq0$ ???

i think the 1 comes from a more logical step?

haha its legal to just do that?
It is in this case, since

$$\frac{1}{2(|l|+1)}<\frac{1}{2|l|}$$

So you're only making the delta smaller. This is certainly allowed.

why not just say that m$\neq0$, l$\neq0$ ???
This is also ok, but then you would have to give a different proof in the case that l=0 or m=0.

It is in this case, since

$$\frac{1}{2(|l|+1)}<\frac{1}{2|l|}$$

So you're only making the delta smaller. This is certainly allowed.

This is also ok, but then you would have to give a different proof in the case that l=0 or m=0.
ohhh gotcha thanks....are you sure though that this 1 cant possibly follow from some earlier step?? i just want to make sure because it is bothering me lol

ohhh gotcha thanks....are you sure though that this 1 cant possibly follow from some earlier step?? i just want to make sure because it is bothering me lol
I looked at Spivak, and this is the only possible reason I see why he would want to add +1.

You should try to see why Spivak's proof works but when I learned this from Spivak it all felt very mysterious. See if you can do the proof yourself:

| (a+$\delta$)(b+$\delta$) - ab | < $\epsilon$

| (a+b)$\delta$ + $\delta$2| < $\epsilon$

can you see why its not too hard to find a $\delta$ that works.

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I'm sorry if this post doesn't help but I never understood this proof unless I could find it out myself.