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## 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:

[itex]\left|f(x)g(x)-lm\right|[/itex]< E

[itex]\leq[/itex][itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E

[itex]\left|f(x)-l\right|[/itex][itex]\leq[/itex]1

[itex]\left|f(x)\right|[/itex][itex]\leq[/itex][itex]\left|l\right|[/itex]+1

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<([itex]\left|l\right|[/itex]+1)[itex]\left|g(x)-m\right|[/itex]<E/2

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

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

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<E/2

ok so now i am guessing that we wanna say that [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E/2 so that E/2 + E/2 = E

so can u just say:

[itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E/2

[itex]\left|f(x)-l\right|[/itex]<E/(2[itex]\left|m\right|[/itex])

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

so:

[itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<[itex]\left|m\right|[/itex](E/(2[itex]\left|m\right|[/itex]))= E/2

so then:

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]< E/2 + E/2 =E

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

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

ok so the following i understand:

[itex]\left|f(x)g(x)-lm\right|[/itex]< E

[itex]\leq[/itex][itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E

[itex]\left|f(x)-l\right|[/itex][itex]\leq[/itex]1

[itex]\left|f(x)\right|[/itex][itex]\leq[/itex][itex]\left|l\right|[/itex]+1

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<([itex]\left|l\right|[/itex]+1)[itex]\left|g(x)-m\right|[/itex]<E/2

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

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

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<E/2

ok so now i am guessing that we wanna say that [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E/2 so that E/2 + E/2 = E

so can u just say:

[itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E/2

[itex]\left|f(x)-l\right|[/itex]<E/(2[itex]\left|m\right|[/itex])

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

so:

[itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<[itex]\left|m\right|[/itex](E/(2[itex]\left|m\right|[/itex]))= E/2

so then:

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]< E/2 + E/2 =E

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

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