- #1
MidgetDwarf
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- Homework Statement
- Prove or disprove that f(x) = { sin (1/x) if x≠0, 1 if x =0} is continuous on [0,infinity] by definition of continuity or one of its equivalent definitions.
- Relevant Equations
- Definition of continuity:
Let f: D → R^q. Let a∈D. We say that f is continuous at a if every neighborhood V of f(a) ∃ a neighborhood U of a (which depends on V) such that if x∈D∩U, then f(x)∈V. (Note that neighborhood of a point in X^n was defined to be a set that contains an open set containing the point x).
The equivalent definitions:
(1)If ∀ε>0, ∃δ>0 such that if ||x-a|| < δ, then ||f(x) - f(a)|| < ∈. (note ||. || is defined to be the standard metric on R^n).
(2) If (Xn) is any sequence of elements of D which converges to a, then the sequence (f(xn)) converges to f(a).
Discontinuity is also defined by (2), ie., showing that the negation of (2) holds, or
The function f is not continuous on a∈D iff there is a sequence (Xn) in D which converges to a, but (f(Xn)) does not converge to f(a).
I know that the function, g(x)= sin(1/x) has infinite oscillations when the values of x get closer and closer to 0. So its limit does not exist (from graphing it). However, the way that we defined f(x), at x=0, f(x)=1, but f(x)= sin(1/x) on (0,infinity).
I have an issue in general showing that piecewise defined functions are continuous/not continuous at a point without resorting to a limit argument.
I am assuming that the way how the function is defined, it fixes the discontinuity issue of g(x) at x=0. ?
I want to show |f(x) - f(0)| <ε, whenever |x-0|<δ.
I know that sin(1/x ≤ |1| ∀x∈R.
Busy work:
So |f(x)- f(o)| = |sin(1/x) - f(0)|≤ |sin(1/x)| + 1 ≤ 1+1=2
but |f(x) -f(o)|= |1 -1|=o.
So my confusion about continuity regarding piecewise defined functions is preventing me from going forward.
I have an issue in general showing that piecewise defined functions are continuous/not continuous at a point without resorting to a limit argument.
I am assuming that the way how the function is defined, it fixes the discontinuity issue of g(x) at x=0. ?
I want to show |f(x) - f(0)| <ε, whenever |x-0|<δ.
I know that sin(1/x ≤ |1| ∀x∈R.
Busy work:
So |f(x)- f(o)| = |sin(1/x) - f(0)|≤ |sin(1/x)| + 1 ≤ 1+1=2
but |f(x) -f(o)|= |1 -1|=o.
So my confusion about continuity regarding piecewise defined functions is preventing me from going forward.