Question

plz solve this n send ​

Answer 1

TO FIND :

${ \bf{ lim_{x \to1}( \frac{1 + cos\pi x}{\pi( {1 - x)}^{2} } ) }}$

ANSWER :

Using L'Hopital's rule

L'Hopital's rule:

Since evaluating the limits of the numerator and denominator would result in an intermediate form

${\sf{ lim_{x \to1}( \frac{ \frac{d}{dx} (1 + cos\pi x)}{ \frac{d}{dx} (\pi {(1 - x)}^{2} )} ) }}$

${ \sf{ lim_{ x\to1}( \frac{ - \pi sin\pi x}{ - 2\pi + 2\pi x} )}}$

Taking common

${ \sf{ lim_{x \to1}( \frac{ - \cancel{ \pi} sin\pi x}{ \cancel{\pi}( - 2 + 2x)} ) }}$

${ \sf{ lim_{x \to1}( - \frac{sin\pi x}{ - 2 + 2x} ) }}$

Again using L'Hopital's rule

$- { \sf{ lim_{x \to1}( \frac{ \frac{d}{dx}( sin\pi x)}{ \frac{d}{dx} ( - 2 + 2x)} ) }}$

${ \sf{ - lim_{x \to1}(\frac{\pi cos\pi x}{2}) }}$

Evaluate the limit

$\to { \sf{ - \frac{\pi cos\pi}{2} }}$

Simply the expression

We know that

π = 180° = 90° + 90°

cos(90° + 90°) = - sin90° = - 1

Substitute

$\to{ \bf{ \frac{ -\pi ( - 1)}{2} = \frac{\pi}{2} }}$

Hence

${ \bf{ lim_{x \to1}( \frac{1 + cos\pi x}{\pi( {1 - x)}^{2} } ) = \frac{\pi}{2} }}$