Question:
Find the value of
\[
\lim_{x\to\infty}
\frac{\sqrt{x}}
{\sqrt{x+\sqrt{x+\sqrt{x}}}}
\]
Find the value of
\[
\lim_{x\to\infty}
\frac{\sqrt{x}}
{\sqrt{x+\sqrt{x+\sqrt{x}}}}
\]
Show Hint
For limits involving nested radicals and \(x\to\infty\), divide by the dominant power of \(x\) to expose small terms tending to zero.
Updated On: Jun 11, 2026
- \(0\)
- \(\sqrt2\)
- \(1\)
- \(\sqrt3\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Factor out \(x\) from the denominator.
\[ L= \lim_{x\to\infty} \frac{\sqrt{x}} {\sqrt{x+\sqrt{x+\sqrt{x}}}}. \] Write \[ L= \lim_{x\to\infty} \frac{1} {\sqrt{1+\dfrac{\sqrt{x+\sqrt{x}}}{x}}}. \]
Step 2: Evaluate the inner fraction.
\[ \frac{\sqrt{x+\sqrt{x}}}{x} = \frac{\sqrt{x}\sqrt{1+\frac1{\sqrt{x}}}}{x} = \frac1{\sqrt{x}} \sqrt{1+\frac1{\sqrt{x}}}. \] As \[ x\to\infty, \] \[ \frac1{\sqrt{x}} \sqrt{1+\frac1{\sqrt{x}}} \to 0. \]
Step 3: Substitute in the limit.
\[ L= \frac1{\sqrt{1+0}} =1. \] Hence \[ \boxed{1}. \]
\[ L= \lim_{x\to\infty} \frac{\sqrt{x}} {\sqrt{x+\sqrt{x+\sqrt{x}}}}. \] Write \[ L= \lim_{x\to\infty} \frac{1} {\sqrt{1+\dfrac{\sqrt{x+\sqrt{x}}}{x}}}. \]
Step 2: Evaluate the inner fraction.
\[ \frac{\sqrt{x+\sqrt{x}}}{x} = \frac{\sqrt{x}\sqrt{1+\frac1{\sqrt{x}}}}{x} = \frac1{\sqrt{x}} \sqrt{1+\frac1{\sqrt{x}}}. \] As \[ x\to\infty, \] \[ \frac1{\sqrt{x}} \sqrt{1+\frac1{\sqrt{x}}} \to 0. \]
Step 3: Substitute in the limit.
\[ L= \frac1{\sqrt{1+0}} =1. \] Hence \[ \boxed{1}. \]
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