InterviewSolution
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InterviewSolution
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`A_(i)=(x-a_(i))/(|x-a_(i)|),i=1,2,...,n," and "a_(1)lta_(2)lta_(3)lt...lta_(n).` If `1lemlen,minN,` then `lim_(xtoa_(m)) (A_(1)A_(2)...A_(n))` |
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Answer» Correct Answer - C We have `A_(i)=(x-a_(i))/(|x-a_(i)|),i=1,2,...n,` and `a_(1)lta_(2)lt...lta_(n-1)lta_(n).` Let x be in the left neighborhood of `a_(m).` Then `x-a_(i)lt0" for "i=m,m+1,…,n` and `x-a_(i)lt0" for "i=1,2,…,m-1` and `A_(i)=(x-a_(i))/(-(x-a_(i)))=-1 " for "i=m,m+1,...,n` and `A_(i)=(x-a_(i))/(x-a_(i))=1 " for "i=1,2,...,m-1` Similarly, if x is in the right neighborhood of `a_(m)`, then `x=a_(i)lt0" for "i=m+1,...,n," and "x-a_(i)lt0" for "i=1,2,...,m.` Therefore, `A_(i)=(x-a_(i))/(-(x-a_(i)))=-1" for "i=m+1,...,n` and `A_(i)=(x-a_(i))/(x-a_(i))=1" for "i=1,2,...,m` Now, Now, `underset(xtoa_(m)^(-))lim(A_(1)A_(2)...A_(n))=(-1)^(n-m+1)` and `underset(xtoa_(m)^(+))lim(A_(1)A_(2)...A_(n))=(-1)^(n-m)` Hence, `underset(xtoa_(m))lim(A_(1)A_(2)...A_(n))` does not exist. |
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