is sin 1 x uniformly continuous

is sin 1 x uniformly continuous Let c 0 xsin 1 x is uniformly continuous on c 8 1 since c 8 1 is compact this implies there exists d 0 such that for every x y c 8 1 x y d implies xsin 1 x ysin 1 y c 4 let x y 0 1 suppose that x y inf d c 8 if x y d 8 xsin 1 x ysin 1

Here is my proof We take the functions g x 1 x and h x sin x now we see that g x is continuous in the open interval 0 0 because it s defined for every x R except in x 0 On the other hand h x is continuous all over reals So it s also continuous in 0 0 then by the fact that the Solution By the definition of uniform continuity for every 0 we must find a 0 such that for all x y in the domain if x y then f x f y Choose an arbitrary 0 First we choose an arbitrary value for Try to Express in Terms of

is sin 1 x uniformly continuous

is sin 1 x uniformly continuous
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Solved Prove That F x Cos x 2 Is Not Uniformly Continuous On R
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Is Sin 1 x A Continuous Function At X 0
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f x 1 x text on a infty where a 0 Answer Add texts here Do not delete this text first Exercise PageIndex 2 Prove that each of the following functions is not uniformly continuous on the given domain f x x 2 on mathbb R f x sin frac 1 x on 0 1 f x ln x on 0 infty Answer F x x sin 1 x in the interval 0 infinity is uniformly continuous using the following definition Given f I R R f is uniformly continuous on I if 0 0 such that x y I x y f x f y

1 Answer sente May 9 2016 The function as given is not continuous at 0 as 0sin 1 0 is not defined However we may make a slight modification to make the function continuous defining f x as f x xsin 1 x if x 0 0 if x 0 We will Notice that the uniform continuity on a set implies the continuity on the same set i e uniform continuity is a stronger continuity Try to prove it Example 1 The function f x 1 1 x2 is uniformly continuous on R Proof Note that for any x u 2 R jf x 1 f u j x2 jx uj jx 1 u2 1 x2 1 u2 uj Hence we estimate

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Solved Show That F x 1 x Is Uniformly Continuous On 1 Chegg
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SOLVED Since Both Cosx And Sin 1 X Are Continuous Functions F x sin
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In general the answer is Example Let f x 1 on 0 1 x Then sn 1 2 0 1 is Cauchy because it converges here n n but f sn 1 n sn 1 2 is not Cauchy because it doesn t converge What is going on here Even though the inputs sn are close together the outputs f sn are very far apart What your professor is saying is that this is an easy way to see that sin 1 x sin 1 x cannot be uniformly continuous Since as Martin R says you can show that any uniformly continuous function on 0 1 0 1 can be extended continuously to 0 1

Continuity y x 3 4 x 1 continuity y frac x 2 x 1 x continuity sqrt 4 x 2 x 2 continuity left frac sin x x x0 right Show More Continuity and Uniform Continuity Below I stands for any one of the intervals a b a b a b a b a 1 a 1 1 b 1 b 1 1 R Let f be a function de ned on an interval I De nition of Continuity on an Interval The function f is continuous on I if it is continuous at every c in I So For every c in I for every

Solved Is The Function F x Sin 1 x Uniformly Continuous Chegg
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Is The Function F x X Sin 1 x Uniformly Continuous Chegg
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is sin 1 x uniformly continuous - F x x sin 1 x in the interval 0 infinity is uniformly continuous using the following definition Given f I R R f is uniformly continuous on I if 0 0 such that x y I x y f x f y