JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 12)
$$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$
exists and equals $${1 \over {2\sqrt 2 }}$$
exists and equals $${1 \over {4\sqrt 2 }}$$
exists and equals $${1 \over {2\sqrt 2 (1 + \sqrt {2)} }}$$
does not exists
Explanation
$$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$
If you put y = 0 at $${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$ it is in $${0 \over 0}$$ form. So we can use L' Hospital's Rule.
= $$\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$$
= $$\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$$
= $${1 \over {4\sqrt 2 }}$$
If you put y = 0 at $${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$ it is in $${0 \over 0}$$ form. So we can use L' Hospital's Rule.
= $$\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$$
= $$\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$$
= $${1 \over {4\sqrt 2 }}$$
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