Question

Solve it.
It is from Limits class 11​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:\lim_{y\to 0}\dfrac{ \sqrt{1 + \sqrt{1 + {y}^{4} } } \: - \: \sqrt{2} }{ {y}^{4} }$

☆ If we substitute directly y = 0, we get indeterminant form

$\tt \:\lim_{y\to 0}\dfrac{ \sqrt{1 + \sqrt{1 + {y}^{4} } } \: - \: \sqrt{2} }{ {y}^{4} } \times \dfrac{ \sqrt{1 + \sqrt{1 + {y}^{4} } } + \sqrt{2} }{ \sqrt{1 + \sqrt{1 + {y}^{4} } } + \sqrt{2} }$

$\tt \: = \lim_{y\to 0}\dfrac{1 + \sqrt{1 + {y}^{4} } - 2}{ {y}^{4} \bigg( \sqrt{1 + \sqrt{1 + {y}^{4} } } + \sqrt{2} \bigg)}$

$\tt \: = \lim_{y\to 0}\dfrac{ \sqrt{1 + {y}^{4} } - 1 }{{y}^{4} \bigg( \sqrt{1 + \sqrt{1 + {y}^{4} } } + \sqrt{2} \bigg)}$

$\tt \: = \lim_{y\to 0}\dfrac{ \sqrt{1 + {y}^{4} } - 1}{{y}^{4} \bigg( \sqrt{1 + \sqrt{1 + {y}^{4} } } + \sqrt{2} \bigg)} \times \dfrac{ \sqrt{1 + {y}^{4} } + 1 }{ \sqrt{1 + {y}^{4} } + 1}$

$\tt \: = \lim_{y\to 0}\dfrac{ \cancel1 + {y}^{4} - \cancel1}{{y}^{4} \bigg( \sqrt{1 + \sqrt{1 + {y}^{4} } } + \sqrt{2} \bigg)\bigg( \sqrt{1 + {y}^{4} } + 1 \bigg)}$

$\tt \: = \lim_{y\to 0}\dfrac{ \cancel{ {y}^{4} }}{ \cancel{{y}^{4}} \bigg( \sqrt{1 + \sqrt{1 + {y}^{4} } } + \sqrt{2} \bigg)\bigg( \sqrt{1 + {y}^{4} } + 1\bigg)}$

$\tt \:  = \dfrac{1}{2 \sqrt{2}(2) }$

$\tt \:  = \dfrac{1}{4 \sqrt{2} } \times \dfrac{ \sqrt{2} }{ \sqrt{2} }$

$\tt \:  = \dfrac{ \sqrt{2} }{8}$

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