use the definition of derivative to prove that
lim x->0 [ln(1+x)]/[x] = 1
[ln (u + h) - ln (u)]/h as h-->0= definition of d/du(ln u) = 1/u
here u = 1
so
1/1 = 1
To prove the given limit using the definition of the derivative, we need to find the derivative of ln(1+x) and then evaluate it at x=0.
The definition of the derivative states that for a function f(x), the derivative f'(x) at a point a is defined as:
f'(a) = lim h->0 [f(a+h) - f(a)] / h
Let's apply this definition to the function f(x) = ln(1+x):
f'(x) = lim h->0 [ln(1+x+h) - ln(1+x)] / h
Now, to simplify this expression, we will use the logarithmic property ln(a) - ln(b) = ln(a/b).
f'(x) = lim h->0 ln[(1+x+h)/(1+x)] / h
Next, we will rewrite the expression inside the logarithm using a common denominator:
f'(x) = lim h->0 ln[[(1+x+h) - (1+x)] / (1+x)] / h
Simplifying the numerator:
f'(x) = lim h->0 ln[h / (1+x)] / h
Applying the property ln(a/b) = ln(a) - ln(b):
f'(x) = lim h->0 [ln(h) - ln(1+x)] / h
As h approaches 0, ln(h) approaches negative infinity, so the term ln(h) will not contribute to the limit.
f'(x) = lim h->0 [-ln(1+x)] / h
Now, let's evaluate this limit as x approaches 0. We can substitute x=0 in the expression:
f'(0) = lim h->0 [-ln(1+0)] / h
f'(0) = lim h->0 [-ln(1)] / h
Since ln(1) = 0, we have:
f'(0) = lim h->0 [0] / h
The expression simplifies to:
f'(0) = lim h->0 0
As h approaches 0, the limit of 0 is equal to 0.
Therefore, f'(0) = 0.
Thus, we have proven that:
lim x->0 [ln(1+x)] / [x] = 1, using the definition of the derivative.