(a+h)^2sin(a+h)-a^2sina/h where limit h tends to zero
asked Mar 16, 2013 in Calculus Answers by anivish (120 points)

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2 Answers

sin(a+h)=sinacosh+cosasinh, so (a+h)^2=a^2+2ah+h^2 and (a+h)^2sin(a+h)-a^2sina=a^2sina(cosh-1)+a^2cosasinh+2ahsinacosh+2ahcosasinh+h^2(sinacosh+cosasinh). As h tends to zero, the first term tends to zero because cosh tends to 1. sinh tends to h when h is small. The second term tends to a^2hcosa and the fourth term,  2ahcosasinh becomes 2ah^2cosa, which tends to zero as h tends to zero. The third term tends to 2ahsina. The h^2 term can be ignored because it is a second degree expression which disappears as h tends to 0. So after dividing through by h, we're left with a^2cosa+2asina, so this is the limit as h approaches zero.

answered Apr 30, 2015 by Rod Top Rated User (588,900 points)

(a+h)^2sin(a+h)- a^2sin a 

= (a^2+2ah + h^2)(sin (a+h) – a^2 ain a 

= a^2( sin (a+h) – sin a)) + (2ah + h^2)(sin (a+h)) 
= a^2 ( sin a cos h + cos a sin h – sin a) + (2ah + h^2)(sin (a+h)) 
= a^2 ( sin a (cos h – 1) + cos a sin h) + (2ah + h^2)(sin (a+h)) 

so given expression deviding by h 

= (a^2 ( sin a (cos h – 1) + cos a sin h) + (2ah + h^2)(sin (a+h))(/h 
= a^2( sin a (cos h-1)/h + cos a sin h/h ) + (2a+h) sin (a+h)) 
= a^2( sin a (- 2 sin ^2 h/2)/h + cos a sin h/h ) + (2a+h) sin (a+h)) 

as h->0 we have sinh/h = 1 and (sin ^2 h/2)/h = 2 ( sin h/2)( sin h/2)/h = 0 and sin (a+h) = sin a giving 

a^2 cos a + 2a sin a that is the limit

answered Jun 29 by viju joseph

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