MCQ
$\mathop {\lim }\limits_{x \to 1} \frac{{1 + \cos \pi \,x}}{{{{\tan }^2}\pi \,x}}$ is equal to
  • A
    $0$
  • $1/2$
  • C
    $1$
  • D
    $2$

Answer

Correct option: B.
$1/2$
b
(b) $\mathop {\lim }\limits_{x \to 1} \frac{{(1 + \cos \pi x)}}{{{{\tan }^2}\pi x}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - \pi \sin \pi x}}{{2\pi \tan \pi x{{\sec }^2}\pi x}}$

[Using  $ L-$ Hospital’s rule]

$ = \mathop {\lim }\limits_{x \to 1} \frac{{ - 1}}{2}{\cos ^3}\pi \,x$$ = \frac{1}{2}$.

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