Hey guys,
I'm trying to prove that f(x)=x^4 is continuous on R, using the e-delta argument, but I can't find a way to break up x^4-a^4 effectively in the forms of x-a. ANybody have any idea here?
I appreciate any hints and help!
Thanks a lot!
Hey guys,
I'm trying to prove that f(x)=x^4 is continuous on R, using the e-delta argument, but I can't find a way to break up x^4-a^4 effectively in the forms of x-a. ANybody have any idea here?
I appreciate any hints and help!
Thanks a lot!
That help? If thats insufficient, maybe prove using similar identities by induction that if x^n is continuous, then x^(n+1) is too? Sorry if this doesn't help, I don't know much about the kind of stuff your trying to do
Fix a>0.
Pick any epsilon > 0.
Choose some positive delta < min { epsilon/(15a^3), a }.
Note that x^4 - a^4 = ( x - a )( x^3 + ax^2 + (a^2)x + a^3 )
When |x-a|<delta, have a-delta < x < a+delta, so, since 0< delta < a, get 0 < x < 2a.
Then | x^3 + ax^2 + (a^2)x + a^3 | = x^3 + ax^2 + (a^2)x + a^3 < (2a)^3 + a(2a)^2 + (a^2)(2a) + a^3 = 15 a^3.
Thus |x-a|<delta implies |x^4 - a^4| = | x - a | * | x^3 + ax^2 + (a^2)x + a^3 | < | x - a | * ( 15 a^3 ) < (delta)( 15 a^3 ) < (epsilon/(15a^3))( 15 a^3 ) = epsilon
Thus 0<|x-a|<delta implies |x^4 - a^4| < epsilon.
Thus lim x->a of x^4 exists and equals a^4.
Thus the function f defined by f(x)=x^4 is continuous at x=a for all a>0.
Since f is even, f is also continuous for x < 0 ( lim x->(-a) f(x) = lim t->a f(-t) = lim t->a f(t) = f(a) = f(-a) ).
Showing f is continuous at 0 is very simple (use delta = the fourth root of epsilon).
Thus f is continuous on all of R.
there is no reason to put any restriction on a.
what we CAN do is restrict how far away from a we let x be. for example, we can require .
then:
|x - a| < 1
but |x| - |a| ≤ |x - a| < 1, so:
|x| < 1 + |a|.
therefore:
. for convenience let us call this number L.
then, letting , we have:
.
a similar (but much more awkward to write down) proof shows that:
is continuous at any real number a, for any natural number n.