Bounding ln(1 + x) by integration

The whole assignment rests on a bound for ln(1 + x) obtained in three lines. You still need to know where it comes from.

The starting bound

For every real number t ≥ 0:

   1 − t  ≤  1/(1 + t)  ≤  1

On the right: 1 + t ≥ 1 > 0, hence 1/(1 + t) ≤ 1.

On the left, work with the difference. It is the only reliable method when an inequality mixes a fraction and a polynomial: cross-multiplying would force you to discuss the sign of the denominator.

   1/(1+t) − (1 − t) = [1 − (1 − t)(1 + t)] / (1 + t)
                     = [1 − (1 − t²)] / (1 + t)
                     = t² / (1 + t)

This quotient is non-negative as soon as t ≥ 0: the left-hand inequality is proved.

Integration preserves order

If u ≤ v on [0 ; x], then the integral of u over [0 ; x] is at most that of v. This is the only tool needed here: the three functions above are continuous on [0 ; x], and 1/(1 + t) has ln(1 + t) as an antiderivative.

   ∫₀ˣ (1 − t) dt  =  x − x²/2
   ∫₀ˣ dt/(1 + t)  =  ln(1 + x)
   ∫₀ˣ 1 dt        =  x

Hence, for every x ≥ 0:

   x − x²/2  ≤  ln(1 + x)  ≤  x

What it gives, and what it does not

Subtracting from x and dividing by x² (x > 0):

   0  ≤  (x − ln(1 + x)) / x²  ≤  1/2

The quantity is indeed bounded, but both bounds are fixed: they do not tighten as x tends to 0. No limit can be extracted from them. It is precisely to fill this gap that the assignment introduces the auxiliary function g, the subject of the next lesson.