$\lim_{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^{4}}}-\sqrt{2}}{y^{4}}=$

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

(B) exists and equals $\frac{1}{2\sqrt{2}(\sqrt{2}+1)}$

(C) exists and equals $\frac{1}{2\sqrt{2}}$

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Limits

lim_{y \to 0} \frac{1 + 1 + y ^{4} - 2}{y ^{4}} =

Answer: (A)

For small

y

, use

(1 + u)^{1/2} \approx 1 + \frac{u}{2}

. First

1 + y^{4} \approx 1 + \frac{y ^{4}}{2}

, so

1 + 1 + y^{4} \approx 2 + \frac{y ^{4}}{2} = 2 1 + \frac{y ^{4}}{4} \approx 2 (1 + \frac{y ^{4}}{8})

. Subtracting

2

leaves

\frac{2 y ^{4}}{8}

, and dividing by

y^{4}

gives the limit

\frac{2}{8} = \frac{1}{4 2}

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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