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發問:
睇唔到既話 :... 顯示更多 睇唔到既話 : http://i834.photobucket.com/albums/zz266/khcluv/equation.png lim [sin(x)-sin(1)] / [x-1] x->1 thanks!! ^^
最佳解答:
def. lim(x->a) [ f(x)-f(a)]/(x-a)= f'(a) f(x)= sin(x), f'(x)= cosx lim(x->1) [ sin(x)- sin(1)]/(x-1)= f'(1)= cos(1) Alternative method. sin(x)- sin(1)= sin[(x-1)+1]-sin(1)= sin(x-1)cos(1)+[cos(x-1)-1]sin(1) while, lim(x->1) sin(x-1)/(x-1)= lim(y->0) siny/y=1, lim(x->1) [cos(x-1)-1]/(x-1)= -lim(y->0) (1-cosy)(1+cosy)/[y(1+cosy)] = - lim(y->0) (siny/y)[siny/(1+cosy)]= 1*0=0 so, lim(x->1) [ sin(x)- sin(1)]/(x-1) =lim(y->0) {siny cos(1) /y + [ cos(y)-1]sin(1)/y } = 1*cos(1)+0*sin(1)= cos(1)
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Find the limit發問:
睇唔到既話 :... 顯示更多 睇唔到既話 : http://i834.photobucket.com/albums/zz266/khcluv/equation.png lim [sin(x)-sin(1)] / [x-1] x->1 thanks!! ^^最佳解答:
def. lim(x->a) [ f(x)-f(a)]/(x-a)= f'(a) f(x)= sin(x), f'(x)= cosx lim(x->1) [ sin(x)- sin(1)]/(x-1)= f'(1)= cos(1) Alternative method. sin(x)- sin(1)= sin[(x-1)+1]-sin(1)= sin(x-1)cos(1)+[cos(x-1)-1]sin(1) while, lim(x->1) sin(x-1)/(x-1)= lim(y->0) siny/y=1, lim(x->1) [cos(x-1)-1]/(x-1)= -lim(y->0) (1-cosy)(1+cosy)/[y(1+cosy)] = - lim(y->0) (siny/y)[siny/(1+cosy)]= 1*0=0 so, lim(x->1) [ sin(x)- sin(1)]/(x-1) =lim(y->0) {siny cos(1) /y + [ cos(y)-1]sin(1)/y } = 1*cos(1)+0*sin(1)= cos(1)
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