# continuity

• Apr 20th 2011, 10:26 AM
continuity
Q1
Let f (xy) =f(x) f(y) for every x, y belongs to R.if f(x) is continuous at any one point x=a, then prove that f(x) is continuous for all x belongs to R-{0}.

Q2
F(x) = 3 if x=0
and [(1+ax+bx^3)/(x^2)]^(1/x) if x is not equal to zero
Find a, b if F(x) is continuous at 0.(looks wrong question )
• Apr 20th 2011, 10:41 AM
TheChaz
Quote:

Originally Posted by ayushdadhwal
Q1
Let f (xy) =f(x) f(y) for every x, y belongs to R.if f(x) is continuous at any one point x=a, then prove that f(x) is continuous for all x belongs to R-{0}.

Q2
F(x) = 3 if x=0
and [(1+ax+bx^3)/(x^2)]^(1/x) if x is not equal to zero
Find a, b if F(x) is continuous at 0.(looks wrong question )

How about showing some work?!
For Q1, you could write out what it means for f to be continuous on the given set. That's what they call "beginning with the end in mind".
Use the given information about a, have a creative (or logical) insight, and string it all together!
• Apr 20th 2011, 10:42 AM
ragnar
I assume that f(xy) means f(x,y), that is to say, the function is not of the product of x and y, but rather it is a function of x and y separately.

If that's true then this first part should be false by taking y/(x-1). It will be continuous, say, at x = 2 but discontinuous at x = 1, which is in R-{0}. Perhaps you've made a mistake typing in the statement of the question.
• Apr 20th 2011, 10:43 AM
TheChaz
Quote:

Originally Posted by ragnar
I assume that f(xy) means f(x,y), that is to say, the function is not of the product of x and y, but rather it is a function of x and y separately.

If that's true then this first part should be false by taking y/(x-1). It will be continuous, say, at x = 2 but discontinuous at x = 1, which is in R-{0}. Perhaps you've made a mistake typing in the statement of the question.

Perhaps you have made a mistake in the interpretation of the question!
Since the function is defined on R (NOT RxR), xy is a product.
• Apr 20th 2011, 10:44 AM
However, I believe that $a\ne 0$ is necessary for this.