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 ExamsLimit, 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
  1. (A–P; B – T; C – Q; D –S)
  2. (A–T; B – P; C – Q; D –S)
  3. (A–P; B – Q; C – T; D –S)
  4. (A–S; B – T; C – Q; D –P)

Answer

(A) (A–P; B – T; C – Q; D –S)

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

More Limit, Continuity and Differentiability questions · Browse all practice questions