Which of the following function (s) not defined at x = 0 has/have… - Vidyadip

MCQ

Which of the following function $(s)$ not defined at $x = 0$ has/have irremovable discontinuity at $x = 0$ ?

A
$f(x) = \frac{1}{{\ln \,\left| x \right|}}$
B
$f(x) = \cos \left( {\frac{{\left| {\sin \,x} \right|}}{x}} \right)$
C
$f(x) = x\sin \frac{\pi }{x}$
✓
$f(x) = \frac{1}{{1 + {2^{\cot \,x}}}}$

✓

Answer

Correct option: D.

$f(x) = \frac{1}{{1 + {2^{\cot \,x}}}}$

d
$\left( 1 \right)\,\,\mathop {Lim}\limits_{x \to {0^ + }} \frac{1}{{\ln \left| x \right|}} = \mathop {Lim}\limits_{x \to {0^ - }} \frac{1}{{\ln \left| x \right|}} = 0$

$\left( 2 \right)\,\,\mathop {Lim}\limits_{x \to {0^ + }} \cos \left( {\frac{{\left| {\sin x} \right|}}{x}} \right)$

$ = \mathop {Lim}\limits_{x \to {0^ - }} \cos \left( {\frac{{\left| {\sin x} \right|}}{x}} \right) = \cos 1$

$\left( 3 \right)\,\,\mathop {Lim}\limits_{x \to {0^ + }} x\sin \left( {\frac{\pi }{x}} \right) = \mathop {Lim}\limits_{x \to {0^ - }} x\sin \left( {\frac{\pi }{x}} \right) = 0$

$\left( 4 \right)\,\,\mathop {Lim}\limits_{x \to {0^ + }} \frac{1}{{1 + {2^{\cot x}}}} = \frac{1}{{1 + \infty }} = 0$

$\mathop {Lim}\limits_{x \to {0^ - }} \frac{1}{{1 + {2^{\cot x}}}} = \frac{1}{{1 + 0}} = 1$

Hence $f\left( x \right) = \frac{1}{{1 + {2^{\cot x}}}}$ has irremovable type of discontinuity at $x=0.$

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