Column–I Column–II (A) Lim _ x x 8 x 8 x = (P) 8 (B) Lim _ x 0 [- ^ 2 ] x^ 2 - [- ^ 2 ]…
Mathematics · JEE Advanced · NTA Exams — Limit, Continuity and Differentiability
Column–I Column–II
(A) \(\operatorname{Lim}_{x \rightarrow \infty} x \cos \frac{\pi}{8 x} \cdot \sin \frac{\pi}{8 x}=\) (P) \(\frac{\pi}{8}\)
(B) \[\operatorname{Lim}_{x \rightarrow 0} \frac{\tan \left[-\pi^{2}\right] x^{2}-\left[-\pi^{2}\right] x^{2}}{\sin ^{2}(x)}=\] (Q) \(\sqrt{2}\)
(C) \(\operatorname{Lim}_{x \rightarrow \infty} \sqrt{\frac{2 x-\sin x+\cos x}{x+\cos ^{2} x+\sin ^{2} x}}=\) (R) \(\frac{8}{\pi}\)
(D) \[\operatorname{Lim}_{x \rightarrow 1}\left(\frac{x^{n}-1}{n(x-1)}\right)^{\frac{1}{x-1}}=\] (S) \(\mathrm{e}^{\frac{\mathrm{n}-1}{2}}\)
(T) 0
The correct matching is
(A) \(\operatorname{Lim}_{x \rightarrow \infty} x \cos \frac{\pi}{8 x} \cdot \sin \frac{\pi}{8 x}=\) (P) \(\frac{\pi}{8}\)
(B) \[\operatorname{Lim}_{x \rightarrow 0} \frac{\tan \left[-\pi^{2}\right] x^{2}-\left[-\pi^{2}\right] x^{2}}{\sin ^{2}(x)}=\] (Q) \(\sqrt{2}\)
(C) \(\operatorname{Lim}_{x \rightarrow \infty} \sqrt{\frac{2 x-\sin x+\cos x}{x+\cos ^{2} x+\sin ^{2} x}}=\) (R) \(\frac{8}{\pi}\)
(D) \[\operatorname{Lim}_{x \rightarrow 1}\left(\frac{x^{n}-1}{n(x-1)}\right)^{\frac{1}{x-1}}=\] (S) \(\mathrm{e}^{\frac{\mathrm{n}-1}{2}}\)
(T) 0
The correct matching is
- (A–P; B – T; C – Q; D –S)
- (A–T; B – P; C – Q; D –S)
- (A–P; B – Q; C – T; D –S)
- (A–S; B – T; C – Q; D –P)
Answer
(A) (A–P; B – T; C – Q; D –S)
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