Q99 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

$\lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}}{x^4}=$

  • Correct answer
    A
    $\frac{1}{4 \sqrt{2}}$
  • B
    $\frac{1}{2 \sqrt{2}}$
  • C
    $\frac{1}{\sqrt{2}}$
  • D
    $\frac{1}{3 \sqrt{2}}$

Solution

$\lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}}{x^4}$ $=\lim _{x \rightarrow 0} \frac{\left(\sqrt{1+x^4}-x^5-x^6-1\right)}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}\right)}$ $=\lim _{x \rightarrow 0} \frac{\left(1+\frac{x^4}{2}-x^5-x^6-1\right)}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}\right)}$ $=\lim _{x \rightarrow 0} \frac{\frac{1}{2}}{(\sqrt{2}+\sqrt{2})}=\frac{1}{4 \sqrt{2}}$

Hence, the correct answer is option: 'A'.

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