![\lim_{x \to \x0} \frac{ sin^{2} \pi cos^{2}x } \sqrt[ \frac{1}{4} ]{x}](https://tex.z-dn.net/?f=+%5Clim_%7Bx+%5Cto+%5Cx0%7D++%5Cfrac%7B+sin%5E%7B2%7D+%5Cpi++cos%5E%7B2%7Dx++%7D+%5Csqrt%5B+%5Cfrac%7B1%7D%7B4%7D+%5D%7Bx%7D+ "\lim_{x \to \x0} \frac{ sin^{2} \pi cos^{2}x } \sqrt[ \frac{1}{4} ]{x}")
[/tex]
kvnmurty:
clarify what the argument of Sin function is. Is cos x inside the sin function. Is square root only for 1/4 ? not for x?
seems to come to 0.. clarify the denominator is it x or xsquare or root x ?
denominator is x to the power 4
yes pi into cosx is inside in sin function
are you sure denominator has x power 4? is it x square? why do u write root of 1/4? canot you write simply 1/2 ? is there something missing?
yes denominator has x^4
ans is pi^2
check it now. done
Answers
Answered by
0
Sin² [ πcos² x ] = sin² [ π(1+cos 2x)/2 ] = sin² [ π/2 + π/2 cos2x]
= cos² [π/2 cos 2x ] = 1/2 * [ 1 + cos ( π cos 2x ) ]
t = π cos 2x as x -> 0, t -> π
x = 1/2 * Cos^-1 ( t/π )
denominator 1/2 x^4
===========================================
Try in some other way: Let us use the Taylor series of expansion for cos x.
cos x = 1 - x² /2 + x^4 /4! - x^6 /6! + .. = 1 - x² y where y = 1/2 - x²/4! + x^4/6! - ...
y -> 1/2 as x->0
So cos² x = (1 + x^4 y² - 2 x² y)
Sin² [ πcos² x ] = sin² [ π + πx^4 y² - 2 π x² y ] sin π+t = - sin t
= sin ² [ πx^4 y² - 2 π x² y ]
Now problem is lim as x-> 0
sin² [πx^4 y² - 2 π x² y ]
= ------------------------------
x^4
sin² [πx^4 y² - 2 π x² y ] [πx^4 y² - 2 π x² y ]²
= ---------------------------------- * ---------------------- multiply & divide by same.
[πx^4 y² - 2 π x² y ]² x^4
= 1² * (πx²y)² [x² y² - 2]² / x^4 = π² y² [ x² y² - 2]²
= as x-> 0 this is = π² y² (-2)² = π² (1/2)² (4) = π²
answer seems to be π²
= cos² [π/2 cos 2x ] = 1/2 * [ 1 + cos ( π cos 2x ) ]
t = π cos 2x as x -> 0, t -> π
x = 1/2 * Cos^-1 ( t/π )
denominator 1/2 x^4
===========================================
Try in some other way: Let us use the Taylor series of expansion for cos x.
cos x = 1 - x² /2 + x^4 /4! - x^6 /6! + .. = 1 - x² y where y = 1/2 - x²/4! + x^4/6! - ...
y -> 1/2 as x->0
So cos² x = (1 + x^4 y² - 2 x² y)
Sin² [ πcos² x ] = sin² [ π + πx^4 y² - 2 π x² y ] sin π+t = - sin t
= sin ² [ πx^4 y² - 2 π x² y ]
Now problem is lim as x-> 0
sin² [πx^4 y² - 2 π x² y ]
= ------------------------------
x^4
sin² [πx^4 y² - 2 π x² y ] [πx^4 y² - 2 π x² y ]²
= ---------------------------------- * ---------------------- multiply & divide by same.
[πx^4 y² - 2 π x² y ]² x^4
= 1² * (πx²y)² [x² y² - 2]² / x^4 = π² y² [ x² y² - 2]²
= as x-> 0 this is = π² y² (-2)² = π² (1/2)² (4) = π²
answer seems to be π²
it means ?
answer is pi square
Similar questions