_ y 0 1 + 1 + y^4 - 2 y^4 = - Vidyadip

MCQ

$\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 }}{{{y^4}}} = $

✓
exists and equals $\frac{1}{{4\sqrt 2 }}$
B
exists and equals $\frac{1}{{2\sqrt 2 \left( {\sqrt 2 + 1} \right)}}$
C
exists and equals $\frac{1}{{2\sqrt 2 }}$
D
does not exist

✓

Answer

Correct option: A.

exists and equals $\frac{1}{{4\sqrt 2 }}$

a
${\left( {1 + x} \right)^n} \cong 1 + nx$ (when $x \to 0$)

So, $\sqrt {1 + {y^4}} = 1 + \frac{{{y^4}}}{2}$

$\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {2 + \frac{{{y^4}}}{2}} - \sqrt 2 }}{{{y^4}}}$

$ = \frac{{\sqrt 2 \left( {1 + \frac{{{y^4}}}{8} - 1} \right)}}{{{y^4}}} = \frac{{\sqrt 2 }}{8} = \frac{1}{{4\sqrt 2 }}$

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