已知f(x)在x=0处连续,且lim(x趋向0)[f(x)/(e^(x/2))-1]=3,求f(0)+f~(0)
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/07 20:32:16
已知f(x)在x=0处连续,且lim(x趋向0)[f(x)/(e^(x/2))-1]=3,求f(0)+f~(0)
![已知f(x)在x=0处连续,且lim(x趋向0)[f(x)/(e^(x/2))-1]=3,求f(0)+f~(0)](/uploads/image/z/1853981-53-1.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%E5%9C%A8x%3D0%E5%A4%84%E8%BF%9E%E7%BB%AD%2C%E4%B8%94lim%EF%BC%88x%E8%B6%8B%E5%90%910%EF%BC%89%5Bf%28x%29%2F%28e%5E%28x%2F2%29%29-1%5D%3D3%2C%E6%B1%82f%280%29%2Bf%7E%280%29)
x趋向0时,[e^(x/2)]-1=0,要使极限存在,则x趋向0时,f(x)=0,即f(0)=0
利用落比塔法则,分子分母求导,得到
lim(x趋向0)[2f'(x)/[e^(x/2)]]=(代入x=0)=2f'(0)=3
则f'(0)=3/2
所以f(0)+f'(0)=3/2
利用落比塔法则,分子分母求导,得到
lim(x趋向0)[2f'(x)/[e^(x/2)]]=(代入x=0)=2f'(0)=3
则f'(0)=3/2
所以f(0)+f'(0)=3/2
已知f(x)在x=0处连续,且lim(x趋向0)[f(x)/(e^(x/2))-1]=3,求f(0)+f~(0)
lim(x趋向于0)f(2x)/x=1,且f(x)连续,则f'(0)=
若已知函数f(x)在x=0处是连续的,lim x趋向0 f(x)+f(-x)/x存在,能否判断出f(x)和f(-x)的极
lim(x趋向于0) f(x)-f(-x)/x 存在 且函数在x=0出连续,为什么f(0)=0?
高数 设f(x)具有连续的二阶导数,且lim[f(x)/x]=0,在x趋向于0的时候.且f’‘(x)=4,求lim[1+
设f(x)在x=0处连续,且lim (f(x)-1)/x=-1,x→0.,求f(0)
求lim(x→0)[(xf'(x))/(2f(x))]^(1/x),其中f(x)在x=0点某邻域内有三阶连续导数,f(0
设函数f(x)有二阶连续导数,且(x->0)lim[f(x)-a]/[e^x^2-1]=0,(x->0)lim[f ‘’
设函数f(x)有二姐连续导数,且(x->0)lim[f(x)-a]/[e^x^2-1]=0,(x->0)lim[f ‘’
已知f(x)连续,f(x)=e^x+∫(0到x)(2+t-x)f(x)dx,求f(x)
设f(x)在0到正无穷大上可导,f(x)>0,limf(x)=1(x趋向正无穷大),若lim[f(x+nx)/f(x)]
设f(x)有二阶连续导数且f'(x)=0,lim(x趋向于0)f''(x)/|x|=1则