If a=limx→0 1+1+x4−2x4 and b=limx→0 sin2x2−1+cosx, then the value of ab3+4 is :

a=limx→01+1+x4−2x4 =limx→01+1+x4−2x4(1+1+x4+2) =limx→0(1+x4)−1x4(1+x4+1).1(1+1+x4+2) =142 b=limx→0(1−cos2x)(2+1+cosx)2−(1+cosx) =limx→0(1+cosx)(2+1+cosx) =(2)(22)=42 So ab3=142×64(22)=32

If a=limx→0 1+1+x4−2x4 and b=limx→0 sin2x2−1+cosx, then the value of ab3+4 is :

a=limx→01+1+x4−2x4 =limx→01+1+x4−2x4(1+1+x4+2) =limx→0(1+x4)−1x4(1+x4+1).1(1+1+x4+2) =142 b=limx→0(1−cos2x)(2+1+cosx)2−(1+cosx) =limx→0(1+cosx)(2+1+cosx) =(2)(22)=42 So ab3=142×64(22)=32

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If $a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}}$ and $b = \lim_{x \to 0} \frac{\sin^{2} x}{\sqrt{2} - \sqrt{1 + \cos x}}$ , then the value of $a b^{3} + 4$ is :

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Detailed Solution

$a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^{4}}} - \sqrt{2}}{x^{4}} = \lim_{x \to 0} \frac{1 + \sqrt{1 + x^{4}} - 2}{x^{4} (\sqrt{1 + \sqrt{1 + x^{4}}} + \sqrt{2})} = \lim_{x \to 0} \frac{(1 + x^{4}) - 1}{x^{4} (\sqrt{1 + x^{4}} + 1)} . \frac{1}{(\sqrt{1 + \sqrt{1 + x^{4}}} + \sqrt{2})} = \frac{1}{4 \sqrt{2}} b = \lim_{x \to 0} \frac{(1 - \cos^{2} x) (\sqrt{2} + \sqrt{1 + \cos x})}{2 - (1 + \cos x)} = \lim_{x \to 0} (1 + \cos x) (\sqrt{2} + \sqrt{1 + \cos x}) = (2) (2 \sqrt{2}) = 4 \sqrt{2} S o a b^{3} = \frac{1}{4 \sqrt{2}} \times 64 (2 \sqrt{2}) = 32$