Question
If $${A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}},\,i = 1,\,2,\,3,.....,\,n$$ and $${a_1} < {a_2} < {a_3}.....{a_n},$$ then $$\mathop {\lim }\limits_{x \to {a_m}} \left( {{A_1}{A_2}.....{A_n}} \right),\,1 \leqslant m \leqslant n$$
A.
is equal to $${\left( { - 1} \right)^m}$$
B.
is equal to $${\left( { - 1} \right)^{m + 1}}$$
C.
is equal to $${\left( { - 1} \right)^{m - 1}}$$
D.
does not exist
Answer :
does not exist
Solution :
$$\eqalign{
& {A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}},\,i = 1,\,2,\,3,.....,\,n \cr
& {a_1} < {a_2} < {a_3} < ..... < {a_n} \cr} $$
If $$x$$ is in the left neighborhood of $${a_1} < {a_2} < ..... < {a_{m - 1}} < x < {a_m} < {a_{m + 1}} < ..... < {a_n}$$
$$\eqalign{
& {A_i} = \frac{{x - {a_i}}}{{x - {a_i}}} = 1,\,i = 1,\,2,.....,\,m - 1\,; \cr
& {A_i} = \frac{{x - {a_i}}}{{\left( {{a_i} - x} \right)}} = - 1,\,i = m,\,m - 1,.....,\,n \cr
& \therefore \,{A_1}{A_2}.....{A_n} = {\left( { - 1} \right)^{n - m + 1}}......({\text{i}}) \cr} $$
If $$x$$ is in the left neighborhood of $${a_m},\,{a_1} < {a_2} < ..... < {a_{m - 1}} < {a_m} < x < {a_{m + 1}} < ..... < {a_n},$$
$$\eqalign{
& {A_i} = \frac{{x - {a_i}}}{{x - {a_i}}} = 1,\,i = 1,\,2,.....,\,n \cr
& \therefore \,{A_1}{A_2}.....{A_n} = {\left( { - 1} \right)^{n - m}}......({\text{ii}}) \cr
& \therefore \,\mathop {\lim }\limits_{x \to a_m^ - } \left( {{A_1}{A_2}.....{A_n}} \right) = {\left( { - 1} \right)^{n - m + 1}} \cr
& {\text{and }}\mathop {\lim }\limits_{x \to a_m^ + } \left( {{A_1}{A_2}.....{A_n}} \right) = {\left( { - 1} \right)^{n - m}} \cr
& \therefore \,L.H.L. \ne R.H.L. \cr
& {\text{Hence, }}\mathop {\lim }\limits_{x \to {a_m}} \left( {{A_1}{A_2}.....{A_n}} \right){\text{ does not exist}}{\text{.}} \cr} $$