# Thread: continuity of composite function

1. ## continuity of composite function

Let f:R-------R be a continuous function defined by f(x+y)=f(x) +f(y) for all x,y belongs to R. if the function is continuous at x=0 then it is continuous at all x.
Sir in solution of this question they do like that.
For f (0) we substitute x=y=0 in the given functional equation relation, to get f (0) ==0 .sir how can we put x=0 and y=0 at our own will. How can we say that x and y are independent.
Again sir if f(x) is a polynomial function (say) then f(x) =x which clearly shows that x and y are not independent, as if i put x=1 then y=f(x) =1 shows that y depends upon x.

2. Let's rewrite the given condition with different letters...
f(a + b) = f(a) + f(b) for ALL/ANY a,b that you choose to be real numbers.
So f(2 + 3) = f(2) + f(3) = f(1) + f(4) = f(1 + 4)
You're confusing independence and the letters x, y.

Since "0" and "0" are both real numbers, we are justified is considering what happens when we let a = 0 and b = 0.
f(0 + 0) = f(0) + f(0)
f(0) = 2f(0)
f(0) = 0
a and b are independent, because we choose them, independently (!), before invoking the function.

In your example, y = f(x). In the given problem, y is an input, not a function value.

3. We use English on this website

4. To show that f is continuous for all real numbers, let b be any real number. You then want to show that $\displaystyle \lim_{x\to b} f(x)= f(b)$.

Do that in two steps:
Since f(x) is continuous at x= a,
\lim_{x\to a}f(x)= a
Let h= x- a so that x= h+ a and f(x)= f(h+ a)= f(h)+ f(a). Of course, as x goes to a, h goes to 0 so
\lim_{x\to a}f(x)= \lim_{h\to 0}f(h)+ f(a)= f(a)
from which it follows that
\lim_{h\to 0} f(h)= 0
which says that f is continuous at x= 0.

To show $\displaystyle \lim_{x\to b} f(x)= f(b)$, now, let h= x- b and repeat the above.

5. Originally Posted by poirot
We use English on this website
We also avoid rudeness on this website.
And what's wrong with the English anyhow? A few verb tenses??