_ x x^3 x^2+1+x^4 -x 2 =

MCQ

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

A
$\frac{1}{\sqrt{2}}$
✓
$\frac{1}{4 \sqrt{2}}$
C
$\frac{-1}{4 \sqrt{2}}$
D
$\frac{-1}{\sqrt{2}}$

✓

Answer

Correct option: B.

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

Let $L=\lim _{x \rightarrow \infty} x^3\left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\}$
$=\lim _{x \rightarrow \infty} x^3\left[x\left\{\sqrt{1+\sqrt{\frac{1}{x^4}+1}}-\sqrt{2}\right\}\right]$
$=\lim _{x \rightarrow \infty} x^4\left\{\sqrt{\left.1+\sqrt{\frac{1}{x^4}+1}-\sqrt{2}\right\}}\right.$
Let $\frac{1}{x^4}=t$, when $x \rightarrow \infty, t \rightarrow 0$
$\therefore L=\lim _{t \rightarrow 0} \frac{\sqrt{1+\sqrt{t+1}}-\sqrt{2}}{t} \quad\left(\frac{0}{0} \text { form }\right)$
Using L'hospital rule, we get
$L=\lim _{t \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+\sqrt{t+1}}} \cdot \frac{1}{2 \sqrt{t+1}}}{1}=\frac{1}{4 \sqrt{2}}$
Hence, the required value is $\frac{1}{4 \sqrt{2}}$.

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