三西安毛子律技大学2例4 设 f(x) 可微且 f(x)>0,(0)=f(0)=1, 求lim f(x)解 f(x)可微必连续，从而limf(x)=f(O)=1,1Lin f(x)lim=ln f(x)=el=elim f(x)* = limex=er>0xx→0x->0In f(x)In[1+ f(x) -1]f(x)-1limlimlimx->0x-0x->0xxxf(x)-f(0)= f'(0)= 1,- limx-0x->0解 例4 设 f x( ) 可微且 f x f f ( ) 0, (0) (0) 1,  = =  求 1 0 lim ( ) . x x f x → f x( ) 可微必连续, 0 lim ( ) (0) 1, x f x f → 从而 = = 1 1 ln ( ) 0 0 lim ( ) lime f x x x x x f x  → → = 0 1 lim ln ( ) e x f x → x  = 0 ln ( ) lim x f x → x = 0 ( ) 1 lim x f x → x − = 0 ( ) (0) lim x 0 f x f → x − = − = = f (0) 1,1 = = e e. 0 ln[1 ( ) 1] lim x f x → x + −