lim x→0 3x/(sinx-x)洛必達(dá)=lim 3/cosx-1->∞2.lim x→0 (1-cosmx)/x^2=lim 2sin^2(mx/2)/x^2=lim 2(mx/2)^2/x^2=m^2/23 lim x→0 sin4x/[(√x+2)-√2]=lim sin4x((√x+2)+√2)/x=8√24 lim h →0 [sin(x+h)-sin(x-...第4題不明=lim h →0 [sin(x+h)-sinx+sinx-sin(x-h)]/h=sin'x+sin'x？？lim h->0 [sin(x+h)-sinx]/h這個(gè)不就是sinx在x處導(dǎo)數(shù)的定義.....lim h->0 [sin(x)-sin(x-h)]/h這個(gè)也是，不過是另一種寫法這個(gè)題你也可以這樣寫lim h →0 [sin(x+h)-sin(x-h)]/h=2lim h →0 [sin(x+h)-sin(x-h)]/2h=2 sin'x=2cosxlim x->1 [(x-1)^2sin(x-1)]=.lim x->1 e^[2sin(x-1)ln(x-1)]=limx->1 e^(2(x-1)ln(x-1)) (x-1=t)=limt->0e^(2t/lnt)=e^0=1 這條的答案是0，可以看看哪錯(cuò)嗎？請(qǐng)幫我回答最後3題1.lim x→0 (1-cos2x)/5x^2+x^32.lim x→∞ [(x+3)(x+1)]^x3. lim x→∏/2 [(2x-∏)^2(tan^2x+1)]thanks..額》。。。lim x→0 (1-cos2x)/5x^2+x^3=lim 2sin^2 / x^2(5+x)=2/5lim x→∞ [(x+3)(x+1)]^x沒寫錯(cuò)？->∞lim x→∏/2 [(2x-∏)^2(tan^2x+1)]（x-π/2=t）limt->0 [(t+π)^2tan^2(t+π/2)+1]=lim [cos^2(t+π)^2/sin^2t]+1=1+1=2lim x→∞ [(x+3)/(x+1)]^xlim x→∞ [(x+3)/(x+1)]^x=lim x→∞ [1+ 2/(x+1)]^x=lim x→∞ [1+ 2/(x+1)]^((x+1)/2）*2x/1+x=lim x→∞e^2x/x+1=e^2