Question

Evaluate \[ \lim_{x\to \infty}x^3\left\{\sqrt{x^2+\sqrt{x^4+1}}-\sqrt2x\right\} \]

• A. \(\sqrt2\)
• B. \(\dfrac{1}{2\sqrt2}\)
• C. \(\dfrac{1}{4\sqrt2}\)
• D. \(\dfrac{1}{\sqrt2}\)

Hint:

For limits involving expressions like \(\sqrt{x^4+1}\) as \(x\to\infty\), factor the highest power of \(x\) and use binomial expansion.

Answer 1

The Correct Option is C

Step 1: Expand for large x.
\(\sqrt{x^4+1} \approx x^2+\dfrac{1}{2x^2}\) for large \(x\). So \(x^2+\sqrt{x^4+1} \approx 2x^2+\dfrac{1}{2x^2}\).

Step 2: Expand the outer square root.
\(\sqrt{2x^2+\dfrac{1}{2x^2}} = \sqrt{2}\,x\sqrt{1+\dfrac{1}{4x^4}} \approx \sqrt{2}\,x\left(1+\dfrac{1}{8x^4}\right) = \sqrt{2}\,x + \dfrac{\sqrt{2}}{8x^3}\).

Step 3: Multiply by \(x^3\).
Expression inside braces \(\approx \dfrac{\sqrt{2}}{8x^3}\). Multiplying by \(x^3\): \(\dfrac{\sqrt{2}}{8} = \dfrac{1}{4\sqrt{2}}\).
\[ \boxed{\dfrac{1}{4\sqrt{2}}} \]