y=cosx/2*cosx/4*`````cosx/2^n
y=(2sinx^n/2*cosx/2*cosx/4*`````cosx/2^n)/2sinx/2^n
y=2^(n-1)cosx/2*cosx/4*`````cosx/2^(n-1)*sin/2^(n-1)/2^nsinx/2^n
y=2^(n-2)cosx/2*cosx/4*`````cosx/2^(n-2)*sin/2^(n-2)/2^nsinx/2^n
…………………………
y=sinx/(2^nsinx/2^n)
因?yàn)閡*sin1/u 當(dāng)u趨于無(wú)窮大時(shí),極限為1
所以cosx/2*cosx/4*`````cosx/2^n n趨于無(wú)窮大!的極限為sinx