Define by and if x is irrational. Prove from the definition that f is discontinuous at 1.
Let . Then there exist .
Does it suffice to say that can't be manipulated to some form of now?
Using the Epsilon Delta definition of continuity, we can show that this function is discontinuous at x=1.
Epsilon-delta definition of continuity:
In this particular problem, c=1.
Let and we shall try to find a
We should find a that satisfies for all the x values may they be rational or irrational. But this is impossible as the two intervals does not have a intersection.
That is for a value does not exist. Therefore the function is discontinuous at x=1.
Hope you understood my reasoning. And may there be any doubt please don't hesitate to ask.
the general strategy to prove f is NOT continuous at some point a, is to find SOME ε > 0, for which we can't find ANY δ > 0,
for which |x - a| < δ implies |f(x) = f(a)| < ε. now let's think about this:
8x and 2x^2 + 8 are both continuous functions. the first is "near" 8 for x "near" 1, and the second is "near" 10, for x "near 1".
this suggests we might be able to use some 0 < ε < 2. let's try it for ε = 1.
in order to know that 8x is within 1 of 8, we need 7 < 8x < 9, that is 7/8 < x < 9/8. that is: if |x - 1| < 1/8,
then |8x - 8| < 1. that's pretty straight-forward. it should be clear that allowing δ > 1/8 won't work,
then for some of the (rational) x's in (1-δ, 1+δ), f(x) will be too far away from f(1) = 8.
now |x - 1| < 1/8 is just saying 7/8 < x < 9/8. so 49/64 < x^2 < 81/64, so 98/64 < 2x^2 < 162/64,
so 610/64 < 2x^2 + 8 < 674/64. but 9 < 610/64, so for all x with |x - 1| < 1/8,
for all of the rational x's, f(x) is within 1 of f(1), and this is true for NONE of the irrational x's with |x - 1| < 1/8.
so no 0 < δ ≤ 1/8 will work (for ε = 1), and no δ > 1/8 will work, either (since SOME x's in (1-δ, 1+δ) will be
make f(x) too far away). so, there's one ε > 0, for which it is impossible to find a suitable δ.