Question:
Evaluate
\[
\lim_{x\to \infty}x^3\left\{\sqrt{x^2+\sqrt{x^4+1}}-\sqrt2x\right\}
\]
Evaluate
\[
\lim_{x\to \infty}x^3\left\{\sqrt{x^2+\sqrt{x^4+1}}-\sqrt2x\right\}
\]
Show Hint
For limits involving expressions like \(\sqrt{x^4+1}\) as \(x\to\infty\), factor the highest power of \(x\) and use binomial expansion.
Updated On: Jun 26, 2026
- \(\sqrt2\)
- \(\dfrac{1}{2\sqrt2}\)
- \(\dfrac{1}{4\sqrt2}\)
- \(\dfrac{1}{\sqrt2}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Simplify the inner square root.
For large \(x\), \[ \sqrt{x^4+1}=x^2\sqrt{1+\frac{1}{x^4}} \] Using \[ \sqrt{1+t}=1+\frac{t}{2}+O(t^2), \] we get \[ \sqrt{x^4+1}=x^2\left(1+\frac{1}{2x^4}+O\left(\frac{1}{x^8}\right)\right) \] \[ \sqrt{x^4+1}=x^2+\frac{1}{2x^2}+O\left(\frac{1}{x^6}\right) \]
Step 2: Substitute in the main radical.
\[ x^2+\sqrt{x^4+1} = x^2+x^2+\frac{1}{2x^2}+O\left(\frac{1}{x^6}\right) \] \[ =2x^2+\frac{1}{2x^2}+O\left(\frac{1}{x^6}\right) \] Thus, \[ \sqrt{x^2+\sqrt{x^4+1}} = \sqrt{2x^2+\frac{1}{2x^2}+O\left(\frac{1}{x^6}\right)} \]
Step 3: Factor \(2x^2\).
\[ \sqrt{2x^2+\frac{1}{2x^2}} = \sqrt{2x^2\left(1+\frac{1}{4x^4}\right)} \] \[ =\sqrt2x\sqrt{1+\frac{1}{4x^4}} \] Using \[ \sqrt{1+t}=1+\frac{t}{2}+O(t^2), \] \[ \sqrt{1+\frac{1}{4x^4}} = 1+\frac{1}{8x^4}+O\left(\frac{1}{x^8}\right) \] Therefore, \[ \sqrt{x^2+\sqrt{x^4+1}} = \sqrt2x+\frac{\sqrt2}{8x^3}+O\left(\frac{1}{x^7}\right) \]
Step 4: Evaluate the limit.
\[ \sqrt{x^2+\sqrt{x^4+1}}-\sqrt2x = \frac{\sqrt2}{8x^3}+O\left(\frac{1}{x^7}\right) \] Multiplying by \(x^3\), \[ x^3\left\{\sqrt{x^2+\sqrt{x^4+1}}-\sqrt2x\right\} = \frac{\sqrt2}{8}+O\left(\frac{1}{x^4}\right) \] Hence, \[ \lim_{x\to \infty}x^3\left\{\sqrt{x^2+\sqrt{x^4+1}}-\sqrt2x\right\} = \frac{\sqrt2}{8} \] \[ =\frac{1}{4\sqrt2} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{1}{4\sqrt2}} \]
For large \(x\), \[ \sqrt{x^4+1}=x^2\sqrt{1+\frac{1}{x^4}} \] Using \[ \sqrt{1+t}=1+\frac{t}{2}+O(t^2), \] we get \[ \sqrt{x^4+1}=x^2\left(1+\frac{1}{2x^4}+O\left(\frac{1}{x^8}\right)\right) \] \[ \sqrt{x^4+1}=x^2+\frac{1}{2x^2}+O\left(\frac{1}{x^6}\right) \]
Step 2: Substitute in the main radical.
\[ x^2+\sqrt{x^4+1} = x^2+x^2+\frac{1}{2x^2}+O\left(\frac{1}{x^6}\right) \] \[ =2x^2+\frac{1}{2x^2}+O\left(\frac{1}{x^6}\right) \] Thus, \[ \sqrt{x^2+\sqrt{x^4+1}} = \sqrt{2x^2+\frac{1}{2x^2}+O\left(\frac{1}{x^6}\right)} \]
Step 3: Factor \(2x^2\).
\[ \sqrt{2x^2+\frac{1}{2x^2}} = \sqrt{2x^2\left(1+\frac{1}{4x^4}\right)} \] \[ =\sqrt2x\sqrt{1+\frac{1}{4x^4}} \] Using \[ \sqrt{1+t}=1+\frac{t}{2}+O(t^2), \] \[ \sqrt{1+\frac{1}{4x^4}} = 1+\frac{1}{8x^4}+O\left(\frac{1}{x^8}\right) \] Therefore, \[ \sqrt{x^2+\sqrt{x^4+1}} = \sqrt2x+\frac{\sqrt2}{8x^3}+O\left(\frac{1}{x^7}\right) \]
Step 4: Evaluate the limit.
\[ \sqrt{x^2+\sqrt{x^4+1}}-\sqrt2x = \frac{\sqrt2}{8x^3}+O\left(\frac{1}{x^7}\right) \] Multiplying by \(x^3\), \[ x^3\left\{\sqrt{x^2+\sqrt{x^4+1}}-\sqrt2x\right\} = \frac{\sqrt2}{8}+O\left(\frac{1}{x^4}\right) \] Hence, \[ \lim_{x\to \infty}x^3\left\{\sqrt{x^2+\sqrt{x^4+1}}-\sqrt2x\right\} = \frac{\sqrt2}{8} \] \[ =\frac{1}{4\sqrt2} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{1}{4\sqrt2}} \]
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