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JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1
Let f(x)=lim⁡x→∞x2n−1x2n+1f\left( x \right) = \lim\limits_{x \to \infty } \frac{{{x^{2n}} - 1}}{{{x^{2n}} + 1}}, then
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Correct option: A
f(x)={1,  ∣x∣>1−1,  ∣x∣<1f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,\;\left| x \right| > 1}\\ { - 1,\;\left| x \right| < 1} \end{array}} \right.
Option A (f(x) = \begin{{cases}} 1, |x| > 1 \\ -1, |x| < 1 \end{{cases}}f(x) = \begin{{cases}} 1, |x| > 1 \\ -1, |x| < 1 \end{{cases}}) is correct because as n approaches infinity, x(2n) approaches 0 when |x| < 1 and approaches infinity when |x| > 1, leading to the given piecewise function.

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lim⁡x→0(1x2−cosec⁡2x)\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right) is equal to lim⁡x→π2cos⁡x(1−sin⁡x)2/3=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{(1-\sin x)^{2 / 3}}= The value of lim⁡h→02{3sin⁡(π6+h)−cos⁡(π6+h)3h(3cos⁡ h−sin⁡ h)}\lim\limits_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\} is lim⁡x→∞x(e1/x−1)\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right) is equal to If f(x)=(x+1)71+x2(x2−x+1)6f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}, then the value of f′(0)f^{\prime}(0) is equal to lim⁡n→∞(1+sin⁡an)n\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n} equals
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