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Question

f(x)=2x+1,a=1,l=3 and ϵ=0.001, then δ>0 satisfying 0<|xa|<δ such that |f(x)l|<ϵ, is

A
0.0005
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B
0.005
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C
0.001
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D
0.0001
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Solution

The correct option is A 0.0005
We have to find δ>0, which satisfies 0<|xa|<δ such that |f(x)l|<ϵ
|f(x)l|<ϵ
ϵ<f(x)l<ϵ
lϵ<f(x)<l+ϵ
3ϵ<2x+1<3+ϵ
ϵ<2x2<ϵ
ϵ2<x1<ϵ2
|x1|<ϵ2
|xa|<ϵ2
δ=ϵ2=0.0012=0.0005
Hence, option A.

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