百度作业网1.lim x→0 3x/(sinx-x)
来源：学生作业帮 编辑：百度作业网作业帮 分类：数学作业 时间：2024/06/30 18:42:39
1.lim x→0 3x/(sinx-x)
2.lim x→0 (1-cosmx)/x^2
3.lim x→0 sin4x/[(√x+2)-√2]
4.lim h →0 [sin(x+h)-sin(x-h)]/h
2.lim x→0 (1-cosmx)/x^2
3.lim x→0 sin4x/[(√x+2)-√2]
4.lim h →0 [sin(x+h)-sin(x-h)]/h
lim x→0 3x/(sinx-x)
洛必达
=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^2/2
3 lim x→0 sin4x/[(√x+2)-√2]
=lim sin4x((√x+2)+√2)/x
=8√2
4 lim h →0 [sin(x+h)-sin(x-h)]/h
=lim h →0 [sin(x+h)-sinx+sinx-sin(x-h)]/h
=sin'x+sin'x
=cosx+cosx
=2cosx
再问： 第4題不明=lim h →0 [sin(x+h)-sinx+sinx-sin(x-h)]/h =sin'x+sin'x ？？
再答： lim h->0 [sin(x+h)-sinx]/h 这个不就是sinx在x处导数的定义..... lim h->0 [sin(x)-sin(x-h)]/h 这个也是，不过是另一种写法 这个题你也可以这样写 lim h →0 [sin(x+h)-sin(x-h)]/h =2lim h →0 [sin(x+h)-sin(x-h)]/2h =2 sin'x =2cosx
再问： lim 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，可以看看哪錯嗎？ 請幫我回答最後3題 1.lim x→0 (1-cos2x)/5x^2+x^3 2.lim x→∞ [(x+3)(x+1)]^x 3. lim x→∏/2 [(2x-∏)^2(tan^2x+1)] thanks..
再答： 额》。。。 lim x→0 (1-cos2x)/5x^2+x^3 =lim 2sin^2 / x^2(5+x) =2/5 lim x→∞ [(x+3)(x+1)]^x没写错？->∞ 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 =2
再问： lim x→∞ [(x+3)/(x+1)]^x
再答： lim 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
洛必达
=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^2/2
3 lim x→0 sin4x/[(√x+2)-√2]
=lim sin4x((√x+2)+√2)/x
=8√2
4 lim h →0 [sin(x+h)-sin(x-h)]/h
=lim h →0 [sin(x+h)-sinx+sinx-sin(x-h)]/h
=sin'x+sin'x
=cosx+cosx
=2cosx
再问： 第4題不明=lim h →0 [sin(x+h)-sinx+sinx-sin(x-h)]/h =sin'x+sin'x ？？
再答： lim h->0 [sin(x+h)-sinx]/h 这个不就是sinx在x处导数的定义..... lim h->0 [sin(x)-sin(x-h)]/h 这个也是，不过是另一种写法 这个题你也可以这样写 lim h →0 [sin(x+h)-sin(x-h)]/h =2lim h →0 [sin(x+h)-sin(x-h)]/2h =2 sin'x =2cosx
再问： lim 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，可以看看哪錯嗎？ 請幫我回答最後3題 1.lim x→0 (1-cos2x)/5x^2+x^3 2.lim x→∞ [(x+3)(x+1)]^x 3. lim x→∏/2 [(2x-∏)^2(tan^2x+1)] thanks..
再答： 额》。。。 lim x→0 (1-cos2x)/5x^2+x^3 =lim 2sin^2 / x^2(5+x) =2/5 lim x→∞ [(x+3)(x+1)]^x没写错？->∞ 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 =2
再问： lim x→∞ [(x+3)/(x+1)]^x
再答： lim 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