Uniform continuity?

Question

Given the function f(x)=1/(x^2)

Show that it is uniformly continuous on (1/2, infinity)
and show that it is not uniformly continuous on (0, 1/2]

Answer 1

The function blows up at 0, and has no significant continuity behavior anywhere else, so the problem lies near 0. The definition for uniform continuity is simply a way of characterizing the property of a function whose changes in behavior depend on changes in input, but not on the particular inputs used. In this case, we can see that violated near 0. Small changes in x near 0 do not correspond to small changes in f in a uniform manner; rather as one approaches 0, one requires larger changes in f per change in x. Thus, the corresponding behavior of the epsilons in this function on the interval (0, 1/2] depend on how close the x's marking the endpoints of the delta are to 0.
This is not a problem on (1/2, infinity) since 1/x^2 continuously approaches 0 on that interval from the bounding value 1/4.
In more precise language, we need to prove that for every e > 0, there exists some d > 0 such that whenever |x - y| < d, we have |f(x) - f(y)| < e for all pairs x, y in the interval (0, 1/2]. Suppose the statement is true. Then choose e > M for any fixed number M. Then there exists some d > 0 such that whenever |x - y| < d, we have a fixed finite bound M on the largest possible change in f between those two points. But this isn't true, since no matter how small we make d, we can always move the endpoints x and y closer to 0 to get some humungo difference e > M. This is an example of proof by contradiction, which is one of the most powerful (though heavily abused) tools in your arsenal.

Answer 2

Nof47d78613943de44cc0c08ec384575e this is not adequate. A function it truly is continuous on a closedf47d78613943de44cc0c08ec384575e bounded period is uniformly continuous on that intervalf47d78613943de44cc0c08ec384575e although a similar isn't actual for the finished genuine line (this is closed yet not bounded). Considerf47d78613943de44cc0c08ec384575e for examplef47d78613943de44cc0c08ec384575e f(x) = sin (e^x). it is continuous everywhere and bounded with techniques from [f47d78613943de44cc0c08ec384575ef47d78613943de44cc0c08ec384575ef47d78613943de44cc0c08ec384575ef47d78613943de44cc0c08ec384575e]-a million,a million yet not uniformly continuous on R on the grounds that as x receives largerf47d78613943de44cc0c08ec384575e further and extra oscillations are squeezed right into a similar area; in case you %. a given ? you could continuously hit upon a ok such that between ok and ok+? there is an entire cycle of the sin function in between them. i imagine a thanks to bypass about proving uniform continuity on your case must be to look at f(x+?) and word in case you could derive an top sure for |f(x+?) -a million,a million f(x)| in words of |?| on my own (i.e. not depending on x). if you should do thatf47d78613943de44cc0c08ec384575e you ought to be able to tutor uniform continuity; for a given ?f47d78613943de44cc0c08ec384575e you in elementary words ought to hit upon a ? that makes your top sure lower than ?.

Answer 3

A function is said to be contious at a point (C) when..
1- the function is defined at the point (c)
2-the limit of the function when x approached to (c) is defind and equal the value of the function at the point c..

F(x) = 1/ x²
Domain = R - {0}

at (1/2 , ∞)
so the function is defined
F(1/2) = 1 / 1/4 = 4
Lim 1 / x² = 4
x→½
so it's contious at (1/2 , ∞)

at (0, 1/2)
the function is not defiend
so it's not contious at (0.1/2)
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