Question:
Evaluate
\[
\lim_{x\to 0}
\frac{\sqrt{11+|x|-6\sqrt{2+|x|}}}
{6-2\sqrt{2+|x|}}
\]
Evaluate
\[
\lim_{x\to 0}
\frac{\sqrt{11+|x|-6\sqrt{2+|x|}}}
{6-2\sqrt{2+|x|}}
\]
Show Hint
Whenever an expression contains nested square roots, try substituting the inner square root by a variable. It often converts the expression into a perfect square.
Updated On: Jun 26, 2026
- \(-1\)
- \(-\dfrac{1}{2}\)
- \(\dfrac{\sqrt{11-6\sqrt{2}}}{3-\sqrt{2}}\)
- \(\dfrac{1}{2}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Put \(t=\sqrt{2+|x|}\).
Then \[ t^2=2+|x| \] and hence \[ |x|=t^2-2. \] The numerator becomes \[ \sqrt{11+t^2-2-6t} = \sqrt{t^2-6t+9}. \] Thus, \[ \sqrt{11+|x|-6\sqrt{2+|x|}} = \sqrt{(t-3)^2}. \] Since \(t\to \sqrt2\lt 3\), \[ \sqrt{(t-3)^2}=3-t. \]
Step 2: Simplify the denominator.
\[ 6-2\sqrt{2+|x|} = 6-2t = 2(3-t). \]
Step 3: Evaluate the limit.
Therefore, \[ \frac{\sqrt{11+|x|-6\sqrt{2+|x|}}} {6-2\sqrt{2+|x|}} = \frac{3-t}{2(3-t)} = \frac12. \] Hence, \[ \lim_{x\to0} \frac{\sqrt{11+|x|-6\sqrt{2+|x|}}} {6-2\sqrt{2+|x|}} = \frac12. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac12} \]
Then \[ t^2=2+|x| \] and hence \[ |x|=t^2-2. \] The numerator becomes \[ \sqrt{11+t^2-2-6t} = \sqrt{t^2-6t+9}. \] Thus, \[ \sqrt{11+|x|-6\sqrt{2+|x|}} = \sqrt{(t-3)^2}. \] Since \(t\to \sqrt2\lt 3\), \[ \sqrt{(t-3)^2}=3-t. \]
Step 2: Simplify the denominator.
\[ 6-2\sqrt{2+|x|} = 6-2t = 2(3-t). \]
Step 3: Evaluate the limit.
Therefore, \[ \frac{\sqrt{11+|x|-6\sqrt{2+|x|}}} {6-2\sqrt{2+|x|}} = \frac{3-t}{2(3-t)} = \frac12. \] Hence, \[ \lim_{x\to0} \frac{\sqrt{11+|x|-6\sqrt{2+|x|}}} {6-2\sqrt{2+|x|}} = \frac12. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac12} \]
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