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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
The function f(x)=lim⁡n→∞x2n−1x2n+1f(x)=\displaystyle\lim_{n\to\infty}\dfrac{x^{2n}-1}{x^{2n}+1} is identical with the function.
Ag(x)=sgn(x−1)g(x)=\text{sgn}(x-1)
Bh(x)=sgn(tan⁡−1x)h(x)=\text{sgn}(\tan^{-1}x)
Cu(x)=sgn(∣x∣−1)u(x)=\text{sgn}(|x|-1)correct
Dv(x)=sgn(cot⁡−1x)v(x)=\text{sgn}(\cot^{-1}x)
Solution
Step 1: Analyse three cases based on ∣x∣|x|: Case ∣x∣<1|x|<1: x2n→0x^{2n}\to 0, so f(x)=0−10+1=−1f(x)=\dfrac{0-1}{0+1}=-1. Case ∣x∣=1|x|=1: x2n=1x^{2n}=1, so f(x)=1−11+1=0f(x)=\dfrac{1-1}{1+1}=0. Case ∣x∣>1|x|>1: x2n→∞x^{2n}\to\infty. Divide numerator and denominator by x2nx^{2n}: f(x)=1−x−2n1+x−2n→11=1f(x)=\dfrac{1-x^{-2n}}{1+x^{-2n}}\to\dfrac{1}{1}=1. Step 2: So:
f(x)={−1,∣x∣<10,∣x∣=11,∣x∣>1f(x)=\begin{cases}-1,& |x|<1\\ 0,& |x|=1\\ 1,& |x|>1\end{cases}
 Step 3: Recognise this as sgn(∣x∣−1)\text{sgn}(|x|-1): - If ∣x∣<1|x|<1: ∣x∣−1<0|x|-1<0, so sgn(∣x∣−1)=−1\text{sgn}(|x|-1)=-1. ✓ - If ∣x∣=1|x|=1: ∣x∣−1=0|x|-1=0, so sgn(∣x∣−1)=0\text{sgn}(|x|-1)=0. ✓ - If ∣x∣>1|x|>1: ∣x∣−1>0|x|-1>0, so sgn(∣x∣−1)=1\text{sgn}(|x|-1)=1. ✓ Correct answer: (3) u(x)=sgn(∣x∣−1)u(x)=\text{sgn}(|x|-1)
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