For x>0 if y=x+x+x+.....∞ then limx→0+x2[−1+x+x+x+......∞][−1+x3+x3+x3+......∞]=

limx→0x2(−1+1+1+4x2)(−1+1+1+4x32) limx→0x2(1+4x−11+4x3−1) limx→04x34x3×1+4x3+11+4x+1=1

For x>0 if y=x+x+x+.....∞ then limx→0+x2[−1+x+x+x+......∞][−1+x3+x3+x3+......∞]=

limx→0x2(−1+1+1+4x2)(−1+1+1+4x32) limx→0x2(1+4x−11+4x3−1) limx→04x34x3×1+4x3+11+4x+1=1

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For $x > 0$ if $y = \sqrt{x + \sqrt{x + \sqrt{x + ..... \infty}}}$ then $\lim_{x \to 0^{+}} \frac{x^{2} [- 1 + \sqrt{x + \sqrt{x + \sqrt{x + ...... \infty}}}]}{[- 1 + \sqrt{x^{3} + \sqrt{x^{3} + \sqrt{x^{3} + ...... \infty}}}]} =$

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$0$

$- 1$

$1$

$\frac{1}{4}$

answer is C.

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Detailed Solution

$\lim_{x \to 0} \frac{x^{2} (- 1 + \frac{1 + \sqrt{1 + 4 x}}{2})}{(- 1 + \frac{1 + \sqrt{1 + 4 x^{3}}}{2})} \lim_{x \to 0} x^{2} (\frac{\sqrt{1 + 4 x} - 1}{\sqrt{1 + 4 x^{3}} - 1}) \lim_{x \to 0} \frac{4 x^{3}}{4 x^{3}} \times \frac{\sqrt{1 + 4 x^{3}} + 1}{\sqrt{1 + 4 x} + 1} = 1$