JEE Mains · Maths · STD 11 - 12. limits
\(\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 }}{{{y^4}}} = \)
- A 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 & Solution
Correct Answer
(A) exists and equals \(\frac{1}{{4\sqrt 2 }}\)
Step-by-step Solution
Detailed explanation
\({\left( {1 + x} \right)^n} \cong 1 + nx\) (when \(x \to 0\)) So, \(\sqrt {1 + {y^4}} = 1 + \frac{{{y^4}}}{2}\) \(\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {2 + \frac{{{y^4}}}{2}} - \sqrt 2 }}{{{y^4}}}\)…
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