`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))`

Answer

C

Explanation

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.