Question

Evaluate:- $\lim_{x \to\frac{ \pi}{2}} \dfrac{1+cos2x}{\left(\pi-2x\right)^2}$

Answer 1

Step-by-step explanation:

How do you evaluate: limx→π/2（tan3x/tanx）without L'Hospital rule?

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limx→π2tan3xtanx

=limx→π2sin3xcos3xsinxcosx

=limx→π2sin3xcos3x×cosxsinx

cos3x

=cos(2x+x)

=cosxcos2x−sin2xsinx

=cosxcos2x−2sin2xcosx

=cosx(cos2x−2sin2x)

∴

limx→π2sin3xcos3x×cosxsinx

=limx→π2sin3xcosx(cos2x−2sin2x)×cosxsinx

=limx→π2sin3xcos2x−2sin2x×1sinx

=limx→π2sin3xcos2xsinx−2sin3x

=sin3π2cosπsinπ2−2sin3π2

=−1(−1)(1)−2(13)

=−1−3

=13

Final answer: 13

Answer 2

GIVEN :–

$\\ \implies\displaystyle \bf \: \lim_{x \to\frac{ \pi}{2}} \dfrac{1+cos2x}{(\pi-2x)^2}$

TO FIND :–

• Value of limit = ?

SOLUTION :–

• Let the function be –

$\\ \implies\displaystyle \bf P= \: \lim_{x \to\frac{ \pi}{2}} \dfrac{1+cos2x}{(\pi-2x)^2}$

• Using L'HOSPITAL rule –

$\\ \implies\displaystyle \bf P= \: \lim_{x \to\frac{ \pi}{2}} \dfrac{0+( - 2) \sin2x}{2(\pi-2x)( - 2)}$

$\\ \implies\displaystyle \bf P= \: \lim_{x \to\frac{ \pi}{2}} \dfrac{ \cancel{( - 2)} \sin2x}{2(\pi-2x)\cancel{( - 2)}}$

$\\ \implies\displaystyle \bf P= \: \lim_{x \to\frac{ \pi}{2}} \dfrac{\sin(2x)}{2(\pi-2x)}$

• Again Using L'HOSPITAL rule –

$\\ \implies\displaystyle \bf P= \: \lim_{x \to\frac{ \pi}{2}} \dfrac{ \cancel2\cos(2x)}{ \cancel2(-2)}$

$\\ \implies\displaystyle \bf P= \: \lim_{x \to\frac{ \pi}{2}} \dfrac{\cos(2x)}{(-2)}$

• Now Apply limits –

$\\ \implies\bf P = \dfrac{\cos \bigg(2 \times \dfrac{\pi}{2} \bigg)}{(-2)}$

$\\ \implies\bf P = \dfrac{\cos (\pi )}{(-2)}$

$\\ \implies\bf P = \dfrac{(- 1)}{(-2)}$

$\\ \implies \large{ \boxed{\bf P = \dfrac{1}{2}}}$