JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 12)
Let $$f(x) = {{1 - x(1 + |1 - x|)} \over {|1 - x|}}\cos \left( {{1 \over {1 - x}}} \right)$$
for x $$ \ne $$ 1. Then
for x $$ \ne $$ 1. Then
$$\mathop {\lim }\limits_{x \to {1^ + }} f(x)$$ = 0
$$\mathop {\lim }\limits_{x \to {1^ - }} f(x)$$ does not exist
$$\mathop {\lim }\limits_{x \to {1^ - }} f(x)$$ = 0
$$\mathop {\lim }\limits_{x \to {1^ + }} f(x)$$ does not exist
Explanation
$$f(x) = {{1 - x(1 + |1 - x|)} \over {|1 - x|}}\cos \left( {{1 \over {1 - x}}} \right)$$
Now, $$\mathop {\lim }\limits_{x \to {1^ - }} f(x)$$
$$ = \mathop {\lim }\limits_{x \to {1^ - }} {{1 - x(1 + 1 - x)} \over {1 - x}}\cos \left( {{1 \over {1 - x}}} \right)$$
$$ = \mathop {\lim }\limits_{x \to {1^ - }} (1 - x)\cos \left( {{1 \over {1 - x}}} \right) = 0$$
and $$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} $$
$${{1 - x(1 + 1 - x)} \over {x - 1}}\cos \left( {{1 \over {1 - x}}} \right)$$
$$ = \mathop {\lim }\limits_{x \to {1^ + }} - (x + 1).cos\left( {{1 \over {x + 1}}} \right)$$, which does not exist.
Now, $$\mathop {\lim }\limits_{x \to {1^ - }} f(x)$$
$$ = \mathop {\lim }\limits_{x \to {1^ - }} {{1 - x(1 + 1 - x)} \over {1 - x}}\cos \left( {{1 \over {1 - x}}} \right)$$
$$ = \mathop {\lim }\limits_{x \to {1^ - }} (1 - x)\cos \left( {{1 \over {1 - x}}} \right) = 0$$
and $$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} $$
$${{1 - x(1 + 1 - x)} \over {x - 1}}\cos \left( {{1 \over {1 - x}}} \right)$$
$$ = \mathop {\lim }\limits_{x \to {1^ + }} - (x + 1).cos\left( {{1 \over {x + 1}}} \right)$$, which does not exist.
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