First slide

Introduction to limits

Question

$A_{i} = \frac{x - a_{i}}{|x - a_{i}|}, i = 1, 2, . . . . n, and a_{1} < a_{2} < a_{3} < . . . . < a_{n} .$

Moderate

Question

If $1 \leq m \leq n, m \in N,$ then the value of $L = \lim_{x \to {a_{m}}^{-}} . . (A_{1} A_{2} . . . . A_{n})$ is

A

always 1
B

always -1
C

${(- 1)}^{n - m + 1}$
D

${(- 1)}^{n - m}$

Solution

$\begin{array}{l} We have A_{i} = \frac{x - a_{i}}{|x - a_{i}|}, i = 1, 2, . ., n, and a_{1} < a_{2} < . . . < a_{n - 1} < a_{n} . \\ Let x be in the left neighborhood of a_{m} . \\ Then x - a_{i} < 0 for i = m, m + 1, . . ., n \\ and x - a_{i} > 0 for i = 1, 2, . . . ., m - 1 \\ and A_{i} = \frac{x - a_{i}}{- (x - a_{i})} = - 1 for i = m, m + 1, . . . ., n \\ A_{i} = \frac{x - a_{i}}{x - a_{i}} = 1 for i = 1, 2, . . . . ., m - 1 \end{array}$
$Sirnilarly, ifx is in the right neighborhood af a_{m}, then \begin{array}{l} x - a_{i} < 0 for i = m + 1, . . ., n and x - a_{i} > 0 for i = 1, 2, . . ., m . \\ Therefore, \\ A_{i} = \frac{x - a_{i}}{- (x - a_{i})} = - 1 for i = m + 1, . . ., n \\ and A_{i} = \frac{x - a_{i}}{x - a_{i}} = 1 for i = 1, 2, . . ., m \\ Now, \lim_{x \to {a_{m}}^{-}} (A_{1} A_{2} . . . A_{n}) = {(- 1)}^{n - m + 1} \\ and \lim_{x \to {a_{m}}^{+}} (A_{1} A_{2} . . . A_{n}) = {(- 1)}^{n - m} \\ Hence, \lim_{x \to a_{m}} (A_{1} A_{2} . . . A_{n}) does not exist . \end{array}$

Question

If $1 \leq m \leq n, m \in N,$ then the value of $R = \lim_{x \to {a_{m}}^{+}} (A_{1} A_{2} . . . A_{n})$ is

A

always 1
B

always -1
C

${(- 1)}^{m + 1}$
D

${(- 1)}^{n - m}$

Solution

$\begin{array}{l} We have A_{i} = \frac{x - a_{i}}{|x - a_{i}|}, i = 1, 2, . ., n, and a_{1} < a_{2} < . . . < a_{n - 1} < a_{n} . \\ Let x be in the left neighborhood of a_{m} . \\ Then x - a_{i} < 0 for i = m, m + 1, . . ., n \\ and x - a_{i} > 0 for i = 1, 2, . . . ., m - 1 \\ and A_{i} = \frac{x - a_{i}}{- (x - a_{i})} = - 1 for i = m, m + 1, . . . ., n \\ A_{i} = \frac{x - a_{i}}{x - a_{i}} = 1 for i = 1, 2, . . . . ., m - 1 \end{array}$
$Sirnilarly, ifx is in the right neighborhood af a_{m}, then \begin{array}{l} x - a_{i} < 0 for i = m + 1, . . ., n and x - a_{i} > 0 for i = 1, 2, . . ., m . \\ Therefore, \\ A_{i} = \frac{x - a_{i}}{- (x - a_{i})} = - 1 for i = m + 1, . . ., n \\ and A_{i} = \frac{x - a_{i}}{x - a_{i}} = 1 for i = 1, 2, . . ., m \\ Now, \lim_{x \to {a_{m}}^{-}} (A_{1} A_{2} . . . A_{n}) = {(- 1)}^{n - m + 1} \\ and \lim_{x \to {a_{m}}^{+}} (A_{1} A_{2} . . . A_{n}) = {(- 1)}^{n - m} \\ Hence, \lim_{x \to a_{m}} (A_{1} A_{2} . . . A_{n}) does not exist . \end{array}$

Question

If $1 \leq m \leq n, m \in N,$ then the value of $R = \lim_{a \to m} (A_{1} A_{2} . . . A_{n})$ is

A

is always equal to -1
B

is always equal to +1
C

does not exist
D

None of these

Solution

$\begin{array}{l} We have A_{i} = \frac{x - a_{i}}{|x - a_{i}|}, i = 1, 2, . ., n, and a_{1} < a_{2} < . . . < a_{n - 1} < a_{n} . \\ Let x be in the left neighborhood of a_{m} . \\ Then x - a_{i} < 0 for i = m, m + 1, . . ., n \\ and x - a_{i} > 0 for i = 1, 2, . . . ., m - 1 \\ and A_{i} = \frac{x - a_{i}}{- (x - a_{i})} = - 1 for i = m, m + 1, . . . ., n \\ A_{i} = \frac{x - a_{i}}{x - a_{i}} = 1 for i = 1, 2, . . . . ., m - 1 \end{array}$
$Sirnilarly, ifx is in the right neighborhood af a_{m}, then \begin{array}{l} x - a_{i} < 0 for i = m + 1, . . ., n and x - a_{i} > 0 for i = 1, 2, . . ., m . \\ Therefore, \\ A_{i} = \frac{x - a_{i}}{- (x - a_{i})} = - 1 for i = m + 1, . . ., n \\ and A_{i} = \frac{x - a_{i}}{x - a_{i}} = 1 for i = 1, 2, . . ., m \\ Now, \lim_{x \to {a_{m}}^{-}} (A_{1} A_{2} . . . A_{n}) = {(- 1)}^{n - m + 1} \\ and \lim_{x \to {a_{m}}^{+}} (A_{1} A_{2} . . . A_{n}) = {(- 1)}^{n - m} \\ Hence, \lim_{x \to a_{m}} (A_{1} A_{2} . . . A_{n}) does not exist . \end{array}$

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