Expert: Alon Mandes - 11/10/2009

Question
QUESTION: Hey I'm not even sure how to attempt this problem without the denominator being zero.

Use the definition of derivative to prove that   

lim(x>0) ln(1+x)/x = 1

Thanks a lot

ANSWER: Here we can use L'Hospital's Rule :

        Ln(x+1)          Ln(x+1)'          1/(x+1)
Lim     ---------- = Lim  --------- = Lim    --------- = 1/(0+1) =1 .
x->0        x        x->0    x'       x->0       1


Alon.

---------- FOLLOW-UP ----------

QUESTION: But I have to use the definition of derivative to prove it

       f(a+h)-f(a)       
Lim     ----------
x->h        h  

Answer
Here we simply can perform the trick by switching h instead of x . It will give the same
result :

         Ln(x+1)          Ln(h+1)          Ln(h+1)-0          Ln(1+h)-Ln(1)
Lim     ---------- = Lim  --------- = Lim    ------------ = Lim  --------------- =
x->0        x        h->0    h        h->0       h          h->0       h

According to the derivative definition, the last term is f'(x) at x=1, where f(x)=Ln(x).
Heres why :
         Ln(x+h)-Ln(x)          Ln(1+h)-Ln(1)
Ln'(x)= Lim -------------- Therefore :  Ln(x){at x=1}=Ln'(1)= Lim  ---------------- .
       h->0      h          h->0        h
Thus,
    Ln(1+h)-Ln(1)
Lim  -------------  = f'(1) . So, we get :
h->0     h


Lim   Ln(x+1)
x->0  ------- = Ln'(1) = 1/x {at x=1} = 1/1 = 1 .
        x
Q.E.D

Alon.