lim(x->1) x^(1/(1-x)) lim(x->0) x^(tanx)的值
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:2024/08/17 06:43:24
lim(x->1) x^(1/(1-x)) lim(x->0) x^(tanx)的值
![lim(x->1) x^(1/(1-x)) lim(x->0) x^(tanx)的值](/uploads/image/z/20199543-15-3.jpg?t=lim%28x-%3E1%29+x%5E%281%2F%EF%BC%881-x%EF%BC%89%29+lim%28x-%3E0%29+x%5E%28tanx%29%E7%9A%84%E5%80%BC)
1.lim(x->1)[x^(1/(1-x))]=?
∵lim(x->1)[lnx/(1-x)]=lim(x->1)[(1/x)/(-1)] (0/0型极限,应用罗比达法则)
=lim(x->1)[(-1)/x]
=-1
∴lim(x->1)[x^(1/(1-x))]=lim(x->1){e^[lnx/(1-x)]}
=e^{lim(x->1)[lnx/(1-x)]}
=e^(-1)
=1/e.
2.lim(x->0)[x^(tanx)]=?
∵lim(x->0)(tanx*lnx)=lim(x->0)[(sinx/x)(1/cosx)(lnx/(1/x))]
=[lim(x->0)(sinx/x)]*[lim(x->0)(1/cosx)]*[lim(x->0)(lnx/(1/x))]
=1*1*[lim(x->0)(lnx/(1/x))] (应用重要极限)
=lim(x->0)(lnx/(1/x))
=lim(x->0)[(1/x/(-1/x²))] (∞/∞型极限,应用罗比达法则)
=lim(x->0)(-x)
=0
∴lim(x->0)[x^(tanx)]=lim(x->0)[e^(tanx*lnx)]
=e^[lim(x->0)(tanx*lnx)]
=e^(0)
=1.
∵lim(x->1)[lnx/(1-x)]=lim(x->1)[(1/x)/(-1)] (0/0型极限,应用罗比达法则)
=lim(x->1)[(-1)/x]
=-1
∴lim(x->1)[x^(1/(1-x))]=lim(x->1){e^[lnx/(1-x)]}
=e^{lim(x->1)[lnx/(1-x)]}
=e^(-1)
=1/e.
2.lim(x->0)[x^(tanx)]=?
∵lim(x->0)(tanx*lnx)=lim(x->0)[(sinx/x)(1/cosx)(lnx/(1/x))]
=[lim(x->0)(sinx/x)]*[lim(x->0)(1/cosx)]*[lim(x->0)(lnx/(1/x))]
=1*1*[lim(x->0)(lnx/(1/x))] (应用重要极限)
=lim(x->0)(lnx/(1/x))
=lim(x->0)[(1/x/(-1/x²))] (∞/∞型极限,应用罗比达法则)
=lim(x->0)(-x)
=0
∴lim(x->0)[x^(tanx)]=lim(x->0)[e^(tanx*lnx)]
=e^[lim(x->0)(tanx*lnx)]
=e^(0)
=1.
lim趋于0((tanx-x)/(x-sinx))^(cotx-1/x)
求极限 lim x趋近于0 [e^(tanx-x) - 1]/(tanx-x)
1、lim(tanx-sinx)/x的立方.x趋向0,2、lim{(2-x)/2}的2/x-1次方.x趋向0,3、lim
lim (tanx-sinx)/x (x→0)的极限; lim (1-cos4x)/xsinx (x→0)的极限
怎样求,当x趋于0时,lim{( tanx)^2/x}.已知的是:当x趋于0时,lim(sinx/x)=1,lim(1-
lim(x-3/x+1)^x lim趋向无穷大
几道极限题!1,lim(x->1)(1-x)tanπ/22,lim(x->0)(tanx-sinx)/x^3
有关高数极限的问题 lim (1/x)^tanx
lim x→0((x+ sinx)/tanx)
lim(x->0)(sinx+tanx)/x
lim->0(tanx-x)\(x-sinx)
求极限:x→0 lim[(1+tanx)^cotx]