The Product of Two Continuous Functions is Continuous
Product of two continous functions
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im2fastfouru, you know, of course, that the product, fg, will be continuous at a as long as the lim, as x goes to a, of f(x)g(x) is equal to f(a)g(a). Therefore, you must prove that, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if |x-a|< [itex]\delta[/itex], then |f(x)g(x)- f(a)g(a)|< [itex]\epsilon[/itex]. Using, of course, the fact that f and g are both continuous at a, that is, that given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if |x-a|< [itex]\delta[/itex] such that |f(x)- f(a)|< [itex]\epsilon[/itex] and similarly for g.
I would recommend you look at |f(x)g(x)- f(x)g(a)+ f(x)g(a)- f(a)g(a)|.
After all, your book asked you to prove this, not us!
write
f(x+h)=f(x)+[f(x+h)-f(x)]
g(x+h)=g(x)+[g(x+h)-g(x)]
note
|f(x+h)-f(x)|<eps1
|g(x+h)-g(x)|<eps2
|f(x+h)-f(x)|,|g(x+h)-g(x)|<eps=max(eps1,eps2)
also recall
|a+b+c|<|a|+|b|+|c|
hint
write
f(x+h)=f(x)+[f(x+h)-f(x)]
g(x+h)=g(x)+[g(x+h)-g(x)]
note
|f(x+h)-f(x)|<eps1
|g(x+h)-g(x)|<eps2
|f(x+h)-f(x)|,|g(x+h)-g(x)|<eps=max(eps1,eps2)
also recall
|a+b+c|<|a|+|b|+|c|
Why on earth would he do so? Halls hints are quite straightforward, he actually almost did all of what was needed to do.>< All he needs to do is show that If(x)I does not get too large as x-->a(or sth), in other words the op just needs to show that f(x) is bounded, which is easy to show since lim(x-->a)f(x)=f(a), so any function who has a limit is bounded at that limit point. all he needs to do is let e-eplylon= 1, and it is easy to show that f(x) is bounded. If(x)I=If(x)-f(a)+f(a)I<=If(x)-f(a)I+If(a)I< 1 +If(a)I
For the OP: do you see how to go now? Like Halls said, you should give some effort on your own next time, if you want to recieve more help. People here won't just give us answers(including me since i often ask for help).
to see this, let the numbers be a+h and b+k and compare the product of ab to that of (a+h)(b+k), when h and k are small.
To solve the problem and because it is fun. I agree, I provided a different (though very slightly) view.Why on earth would he do so? Halls hints are quite straightforward.
theorem f is continuous if and only iflurflurf may has misread "continuous" as "differentiable".
lim_h->0 (f(x+h)-f(x))=0
where (for picking nits) we accept as implicit f(x) exist and limit_h->0 f(x+h) exist
A useful frameworkthe point (of continuity) is simply that if two numbers are respectively near two other numbers, then the products are also near each other.to see this, let the numbers be a+h and b+k and compare the product of ab to that of (a+h)(b+k), when h and k are small.
we desire to show that
if
|f(x+h)-f(x)|<eps1
|g(x+h)-g(x)|<eps2
then
|f(x+h)g(x+h)-f(x)g(x)|<eps
write
f(x+h)=f(x)+[f(x+h)-f(x)]
g(x+h)=g(x)+[g(x+h)-g(x)]
thus
|f(x+h)g(x+h)-f(x)g(x)|=|[f(x+h)-f(x)]g(x)+f(x)[g(x+h)-g(x)]+[f(x+h)-f(x)][g(x+h)-g(x)]|
<|[f(x+h)-f(x)||g(x)|+|f(x)||g(x+h)-g(x)|+|f(x+h)-f(x)||g(x+h)-g(x)]|
<eps1|g(x)|+|f(x)|eps2+eps1*eps2
letting
eps1=(eps/3)/|g(x)| if |g(x)|>0
eps2=(eps/3)/max(eps1,|f(x)|) if |f(x)|>0
we easily see the result is true
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Source: https://www.physicsforums.com/threads/product-of-two-continous-functions.214450/
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