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Question: $\underset{y\to\infty}{Lt} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}=$...

Lty→∞1+1+y4−2y4=\underset{y\to\infty}{Lt} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}=

A

424\sqrt{2}

B

142\frac{1}{4\sqrt{2}}

C

2\sqrt{2}

D

0

Answer

0

Explanation

Solution

Let x=y4x = y^4. As y→∞y \to \infty, x→∞x \to \infty. The limit becomes Ltx→∞1+1+x−2x\underset{x\to\infty}{Lt} \frac{\sqrt{1+\sqrt{1+x}}-\sqrt{2}}{x}. Using rationalization: Ltx→∞(1+1+x−2)(1+1+x+2)x(1+1+x+2)\underset{x\to\infty}{Lt} \frac{(\sqrt{1+\sqrt{1+x}}-\sqrt{2})(\sqrt{1+\sqrt{1+x}}+\sqrt{2})}{x(\sqrt{1+\sqrt{1+x}}+\sqrt{2})} =Ltx→∞1+1+x−2x(1+1+x+2)= \underset{x\to\infty}{Lt} \frac{1+\sqrt{1+x}-2}{x(\sqrt{1+\sqrt{1+x}}+\sqrt{2})} =Ltx→∞1+x−1x(1+1+x+2)= \underset{x\to\infty}{Lt} \frac{\sqrt{1+x}-1}{x(\sqrt{1+\sqrt{1+x}}+\sqrt{2})} For large xx, 1+x≈x\sqrt{1+x} \approx \sqrt{x} and 1+1+x≈x=x1/4\sqrt{1+\sqrt{1+x}} \approx \sqrt{\sqrt{x}} = x^{1/4}. The expression becomes approximately Ltx→∞xx(x1/4+2)≈Ltx→∞x1/2x5/4=Ltx→∞x−3/4=0\underset{x\to\infty}{Lt} \frac{\sqrt{x}}{x(x^{1/4}+\sqrt{2})} \approx \underset{x\to\infty}{Lt} \frac{x^{1/2}}{x^{5/4}} = \underset{x\to\infty}{Lt} x^{-3/4} = 0.