Q99 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

$\lim _{x \rightarrow 0} x^3\left\{\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2} x\right\}=$

  • Correct answer
    A
    $0$
  • B
    $\frac{1}{2 \sqrt{2}}$
  • C
    $\frac{1}{4 \sqrt{2}}$
  • D
    $\frac{1}{\sqrt{2}}$

Solution

$\begin{aligned} & \lim _{x \rightarrow 0} x^3\left(\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2} x\right) \\ &= \lim _{x \rightarrow 0} \frac{\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right.}{\left(\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x\right.} \\ & \quad \times x^3\left(\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2} x\right) \\ &= \lim _{x \rightarrow 0}\left(x^3\right)\left(\sqrt{x^4+1}-x^2\right) \\ & \quad \times \lim _{x \rightarrow 0} \frac{1}{\sqrt{x^2+\sqrt{x^4+1}}+\sqrt{2} x} \\ &= \lim _{x \rightarrow 0} \frac{\left(x^3\right)\left(\sqrt{x^4+1}-x^2\right)\left(\sqrt{x^4+1}+x^2\right.}{\sqrt{x^4+1}+x^2} \\ &= \lim _{x \rightarrow 0} \frac{x^3}{\sqrt{x^4+1}+x^2}=0\end{aligned}$

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

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