Prove limit theorem f(x)g(x)=lm

[jessjolt2]

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 don't 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|$$\left|g(x)-m\right|$<E/2ok so now i am guessing that we want to say that $\left|m\right|$$\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 can't 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?

 

[lippyka]

Yes, this is correct. The reason why Spivak is using the expression E/(2(|m|+1)) is because m is a constant and thus it always remains the same. So, if m is very large, it might make the expression E/(2(|m|+1)) smaller than E/(2|m|). This would give us a tighter bound on the value of \left|f(x)-l\right|, which would mean that the proof works better when we have larger values of m.