IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

juana

$\mathop{\lim}\limits_{{x} \to {0}} \frac{cos(m+2)x-cosmx}{cos(m+4)x-cos(m+2)x}$

I got 1 but the answer is m+1/m+3

sagarvaze

cos(m+2)x - cosmx / cos(m+4)x-cos(m+2)x

= sin((m+1)x)sin(x)/sin((m+3)x)sin(x)

now as x----->0 sin(kx) = kx

so answer is m+1/m+3

studyid

$\mathop{\lim}\limits_{{x} \to {0}}\frac{cos(m+2)x-cosmx}{cos(m+4)x-cos(m+2)x}$

$=\mathop{\lim}\limits_{{x} \to {}}\frac{-sin(m+1)x\:sinx}{-sin(m+3)x \:sinx}$

$=\mathop{\lim}\limits_{{x} \to {0}}\frac{sin(m+1)x}{(m+1)x}\frac{(m+3)x}{sin(m+3)x}\frac{(m+1)x}{(m+3)x}$

Applying the limits the answer is :

$\frac{m+1}{m+3}$

nunknown91

= $\mathop{\lim}\limits_{{x} \to {0}} \frac{-2sin(m+1)x.sinx}{-2sin(m+3)x.sinx}$

= $\frac{m+1}{m+3}$

i hope u got it...

eshaan

see its simple

apply cosA-cosB formula to get

cancel out sinx and divide and multiply first by m+1 and then by m+3 to get

m+1/m+3

sarvpriye

Just Apply the L-Hospital rule twice

by applying it once, u get

Lim_{x-> 0}( - (m+2) sin(m+2)x + msinx)/ (- (m+4)sin(m+4)x + (m+2) sin(m+2)x )

again applying L-Hospital,

Lim_{x-> 0}(-(m+2)²cos(m+2)x + m²cosmx )/ ( -(m+4)²cos(m+4)x + (m+2)²cos(m+2)x)

Putting limits, we get

(m²- (m+2)²)/( (m+2)² - (m+4)² )

=> (m+1)/ ( m+3)

Rate me..........

juana

Ok thanks everyone