Limits, Continuity & Differentiability
Nested Radical Limit and Trigonometric Limit
nta_pyq_2024_jan
Grade 12
Question:
If $a=\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $b=\displaystyle\lim_{x\to0}\dfrac{\sin^2x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $ab^3$ is:
Step-by-Step Solution
Key Concept: For $a$: rationalize $\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}$ by multiplying by conjugate, then simplify $\sqrt{1+x^4}-1\sim x^4/2$, giving $a=\frac{1}{4\sqrt{2}}$. For $b$: rationalize the denominator $\sqrt{2}-\sqrt{1+\cos x}=(1-\cos x)/(\sqrt{2}+\sqrt{1+\cos x})$, use $\sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x)$, get $b=4\sqrt{2}$.
$a=\frac{1}{4\sqrt{2}}$, $b=4\sqrt{2}$. $ab^3=\frac{1}{4\sqrt{2}}\times128\sqrt{2}=32$.
Correct Answer: 2