Can the Product, Sum, and Quotient of Continuous Functions be Continuous?

[Benny]

Hi, I would like to know if I can say that products, sums, and quotients of continuous functions are continuous. From what I can tell, what I've asked is the same as asking if the product, sums, quotients of limits 'work' and of course they do.

For example if lim(x->a)g(x) = c and lim(x->a)h(x) = d then lim(x->a)g(x)h(x) = cd. If I write c = g(a) and d = h(a) then lim(x->a)g(x)h(x) = g(a)h(a) and I've got continuity of g(x)h(x) at x = a?

The answer looks obvious but I'd like to make sure. Any input would be good thanks.

 

[topsquark]

Hi, I would like to know if I can say that products, sums, and quotients of continuous functions are continuous. From what I can tell, what I've asked is the same as asking if the product, sums, quotients of limits 'work' and of course they do.

For example if lim(x->a)g(x) = c and lim(x->a)h(x) = d then lim(x->a)g(x)h(x) = cd. If I write c = g(a) and d = h(a) then lim(x->a)g(x)h(x) = g(a)h(a) and I've got continuity of g(x)h(x) at x = a?

The answer looks obvious but I'd like to make sure. Any input would be good thanks.

It looks good, but you need to be slightly more careful than this about the quotients.

Another (possibly better, but more advanced) way to approach this would be in the form of an "epsilon-delta" proof. Given a function f(x), continuity of f(x) means that given any \epsilon>0 there exists \delta>0 such that: |x-x_0|<\delta implies that |f(x)-f(x_0)|<\epsilon.

-Dan

 

[VietDao29]

...\epsilon>0 there exists \delta>0 such that: |x-x_0|<\delta...

Hmm, you forgot a zero in the inequality.
It should read:
0 < |x - x_0| < \delta \Rightarrow |f(x) - L| < \varepsilon

 

[Benny]

Ok thanks for the suggestions but if the denominator of the quotient is non-zero over the domain being considered then the quotient of two continuous functions is continuous isn't it? Isn't this just the quotient of limits but with b, c or whatever value of the limit is, replaced with f(something)? So instead of lim(x->a)d(x) = b, you just write d(a) in place of b.

 

[devious_]

Hmm, you forgot a zero in the inequality.
It should read:
0 < |x - x_0| < \delta \Rightarrow |f(x) - L| < \varepsilon

For continuity you actually don't need the zero.

 

[devious_]

Ok thanks for the suggestions but if the denominator of the quotient is non-zero over the domain being considered then the quotient of two continuous functions is continuous isn't it? Isn't this just the quotient of limits but with b, c or whatever value of the limit is, replaced with f(something)? So instead of lim(x->a)d(x) = b, you just write d(a) in place of b.

Yes, that's right.