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lim⁡y→0\mspace2mu1+1+y4−2y4\lim_{y \rightarrow 0}\mspace{2mu}\frac{\sqrt{1 + \sqrt{1 + y^{4}}} - \sqrt{2}}{y^{4}}

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Using rationalization
lim⁡y→0\mspace2mu1+1+y4−2y4×1+1+y4+21+1+y4+2\lim_{y \rightarrow 0}\mspace{2mu}\frac{\sqrt{1 + \sqrt{1 + y^{4}}} - \sqrt{2}}{y^{4}} \times \frac{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}}{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}}
=lim⁡y→0\mspace2mu1+y4−1y4×11+1+y4+2×1+y4+11+y4+1= \lim_{y \rightarrow 0}\mspace{2mu}\frac{\sqrt{1 + y^{4}} - \sqrt{1}}{y^{4}} \times \frac{1}{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}} \times \frac{\sqrt{1 + y^{4}} + 1}{\sqrt{1 + y^{4}} + 1}
=lim⁡y→0\mspace2mu11+1+y4+2×11+y4+1= \lim_{y \rightarrow 0}\mspace{2mu}\frac{1}{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}} \times \frac{1}{\sqrt{1 + y^{4}} + 1}
by putting value of limit
=122×12=142= \frac{1}{2\sqrt{2}} \times \frac{1}{2} = \frac{1}{4\sqrt{2}}
Using L.H rule (L.H.)
lim⁡y→0\mspace2mu121+1+y4⋅4y321+y44y3=142\lim_{y \rightarrow 0}\mspace{2mu}\frac{\frac{1}{2\sqrt{1 + \sqrt{1 + y^{4}}}} \cdot \frac{4y^{3}}{2\sqrt{1 + y^{4}}}}{4y^{3}} = \frac{1}{4\sqrt{2}}

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx

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