Question

Plz solve both of them
No useless answer plz​

Answer 1

QUESTION 1:

$\displaystyle\lim_{x \to \pi/4} \dfrac{4 \sqrt{2} - ( { \cos x + \sin x)}^{5} }{1 - \sin2x}$

ANSWER 1:

5√2

NEED TO USE L' HOSPITAL RULE:

$\displaystyle\lim_{x \to \pi/4} \dfrac{4 \sqrt{2} - ( { \cos x + \sin x)}^{5} }{1 - \sin2x} \\ \\ \\ \dfrac{4 \sqrt{2} - ( { \cos \dfrac{ \pi}{4} + \sin \dfrac{ \pi}{4} )}^{5} }{1 - \sin 2\dfrac{ \pi}{4} } \\ \\ \\ \dfrac{4 \sqrt{2} - {( \dfrac{1}{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} } )}^{5} }{1 - \sin \dfrac{ \pi}{2} }\\ \\ \\ \dfrac{4 \sqrt{2} - {( \dfrac{2}{ \sqrt{2} } )}^{5} }{1 -1 }\\ \\ \\ \dfrac{4 \sqrt{2} - {( \sqrt{2}) }^{5} }{1 -1 } \\ \\ \\ \dfrac{4 \sqrt{2} - 4 \sqrt{2} }{1 -1 } \\ \\ \\ \dfrac{0}{0} = \infty$

HENCE WE NEED TO USE L' HOSPITAL RULE.

FORMULAE USED:

sin (π/4) = cos (π/4) = 1/√2

sin (π/2) = 1

cos 2x = cos² x - sin² x

(A + B)(A - B) = A² - B²

d/dx ( sin 2x) = 2 cos 2x

d/dx(2x) = 2

d/dx (cos x) = - sin x

d/dx ( sin x) = cos x

d/dx(constant) = 0

QUESTION 2:

$\displaystyle\lim_{x \to 4} \frac{ {( \cos \alpha)}^{x} - {( \sin \alpha)}^{x} - \cos2 \alpha}{x - 4}$

ANSWER 2:

${ \cos}^{4} \alpha \log( \cos \alpha ) - { \sin}^{4} \alpha\log( \sin \alpha )$

NEED TO USE L' HOSPITAL RULE:

$\displaystyle\lim_{x \to 4} \frac{ {( \cos \alpha)}^{x} - {( \sin \alpha)}^{x} - \cos2 \alpha}{x - 4}\\ \\ \\ \dfrac{ {( \cos \alpha)}^{4} - {( \sin \alpha)}^{4} - \cos2 \alpha}{4 - 4}\\ \\ \\\dfrac{ {\cos}^{2} \alpha + { \sin}^{2} \alpha{(\cos}^{2} \alpha - { \sin}^{2} \alpha)- ({ \cos}^{2} \alpha - { \sin}^{2} \alpha) }{4 - 4}\\ \\ \\ \dfrac{{(\cos}^{2} \alpha - { \sin}^{2} \alpha)- ({ \cos}^{2} \alpha - { \sin}^{2} \alpha) }{4 - 4} \\ \\ \\ \dfrac{0}{0} = \infty$

HENCE WE NEED TO USE L' HOSPITAL RULE.

FORMULAE USED:

A⁴ - B⁴ = (A² + B²)(A² - B²)

cos 2x = cos² x - sin² x

cos² x + sin² x = 1

$\dfrac{d}{dx} ( {a}^{x} ) = {a}^{x} \log a$

d/dx(constant) = 0

d/dx (x) = 1

NOTE : REFER ATTACHMENT FOR CALCULATION.